第10章三角法のさらなる応用Further Applications of Trigonometry

10.1 Non-right Triangles: Law of Sines10.1 一般の三角形と正弦定理

Learning Objectives 学習目標

In this section, you will:

この節を終えると、次のことができるようになる。

• Use the Law of Sines to solve oblique triangles.• Find the area of an oblique triangle using the sine function.• Solve applied problems using the Law of Sines.

・正弦定理を使って一般の三角形の辺や角を求める。・正弦関数を使って一般の三角形の面積を求める。・正弦定理を使って応用問題を解く。

To ensure the safety of over 5,000 U.S. aircraft flying simultaneously during peak times, air traffic controllers monitor and communicate with them after receiving data from the robust radar beacon system. Suppose two radar stations located 20 miles apart each detect an aircraft between them. The angle of elevation measured by the first station is 35 degrees, whereas the angle of elevation measured by the second station is 15 degrees. How can we determine the altitude of the aircraft? We see in Figure 1 that the triangle formed by the aircraft and the two stations is not a right triangle, so we cannot use what we know about right triangles. In this section, we will find out how to solve problems involving non-right triangles.

米国では混雑時に5,000機を超える航空機が同時に飛行する。航空管制官は、レーダー・ビーコンシステムから得たデータを使って航空機の位置を把握し、通信によって安全を支えている。ここで、20マイル離れた2つのレーダー局が、その間にある航空機を捉えたとしよう。一方で測った仰角が35度、もう一方では15度なら、航空機の高度をどう求めればよいだろうか。図1のように、航空機と2つの局を結ぶ三角形は直角三角形ではない。これまでの直角三角形の公式を、そのまま適用することはできない。この節では、こうした一般の三角形を含む問題の解き方を学ぶ。

A diagram of a triangle where the vertices are the first ground station, the second ground station, and the airplane in the air between them. The angle between the first ground station and the plane is 15 degrees, and the angle between the second station and the airplane is 35 degrees. The side between the two stations is of length 20 miles. There is a dotted line perpendicular to the ground side connecting the airplane vertex with the ground - an altitude line.

Using the Law of Sines to Solve Oblique Triangles正弦定理を使って一般の三角形の辺や角を求める

In any triangle, we can draw an altitude, a perpendicular line from one vertex to the opposite side, forming two right triangles. It would be preferable, however, to have methods that we can apply directly to non-right triangles without first having to create right triangles.

どの三角形でも、一つの頂点から向かいの辺へ垂線、すなわち高さを引けば、二つの直角三角形ができる。だが、まず直角三角形を作らずに、直角でない三角形に直接当てはめられる方法があるほうが望ましい。

Any triangle that is not a right triangle is an oblique triangle. Solving an oblique triangle means finding the measurements of all three angles and all three sides. To do so, we need to start with at least three of these values, including at least one of the sides. We will investigate three possible oblique triangle problem situations:

直角三角形でない三角形はどれも斜三角形である。一般の三角形の辺や角を求めるとは、三つの角すべてと三つの辺すべての大きさを求めることである。そのためには、少なくとも一辺を含めて、この値のうち少なくとも三つから始める必要がある。斜三角形の問題としてありうる三つの状況を調べる。

1. ASA (angle-side-angle) We know the measurements of two angles and the included side. See Figure 2. Figure 22. AAS (angle-angle-side) We know the measurements of two angles and a side that is not between the known angles. See Figure 3. Figure 33. SSA (side-side-angle) We know the measurements of two sides and an angle that is not between the known sides. See Figure 4. Figure 4

1. ASA(角-辺-角) 二つの角と、そのあいだの辺の大きさが分かっている。図2を見よ。図22. AAS(角-角-辺) 二つの角と、その分かっている角のあいだにない辺の大きさが分かっている。図3を見よ。図33. SSA(辺-辺-角) 二つの辺と、その分かっている辺のあいだにない角の大きさが分かっている。図4を見よ。図4

Knowing how to approach each of these situations enables us to solve oblique triangles without having to drop a perpendicular to form two right triangles. Instead, we can use the fact that the ratio of the measurement of one of the angles to the length of its opposite side will be equal to the other two ratios of angle measure to opposite side. Let’s see how this statement is derived by considering the triangle shown in Figure 5.

この状況のそれぞれへの近づき方を知っていれば、垂線を下ろして二つの直角三角形を作らずに斜三角形が解ける。代わりに、一つの角の大きさとその対辺の長さの比が、ほかの二つの角の大きさと対辺の比に等しいという事実が使える。図5に示す三角形を考えて、この主張がどう導かれるかを見よう。

An oblique triangle consisting of sides a, b, and c, and angles alpha, beta, and gamma. Side c is opposide angle gamma and is the horizontal base of the triangle. Side b is opposite angle beta, and side a is opposite angle alpha. There is a dotted perpendicular line - an altitude - from the gamma angle to the horizontal base c.

Using the right triangle relationships, we know that $\sin α = \frac{h}{b}$ and $\sin β = \frac{h}{a} .$ Solving both equations for $h$ gives two different expressions for $h .$

直角三角形の関係を使えば、 $\sin α = \frac{h}{b}$ と $\sin β = \frac{h}{a}$ が分かる。どちらの式も $h$ について解けば、 $h$ の違う二つの式が得られる。

h = b \sin α and h = a \sin β

We then set the expressions equal to each other.

そしてこの式どうしを等しいとおく。

\begin{array}{l} b \sin α = a \sin β \\ (\frac{1}{a b}) (b \sin α) = (a \sin β) (\frac{1}{a b}) \begin{matrix}  \end{matrix} & Multiply both sides by \frac{1}{a b} . \\ \frac{\sin α}{a} = \frac{\sin β}{b} \end{array}

Similarly, we can compare the other ratios.

同じように、ほかの比も比べられる。

\frac{\sin α}{a} = \frac{\sin γ}{c} and \frac{\sin β}{b} = \frac{\sin γ}{c}

Collectively, these relationships are called the Law of Sines.

これらの関係を合わせて正弦定理と呼ぶ。

\frac{\sin α}{a} = \frac{\sin β}{b} = \frac{\sin γ}{c}

Note the standard way of labeling triangles: angle $α$ (alpha) is opposite side $a;$ angle $β$ (beta) is opposite side $b;$ and angle $γ$ (gamma) is opposite side $c .$ See Figure 6.

三角形に名前を付ける標準のやり方に注意しよう。角 $α$ (アルファ)は辺 $a;$ の向かい、角 $β$ (ベータ)は辺 $b;$ の向かい、角 $γ$ (ガンマ)は辺 $c$ の向かいである。図6を見よ。

While calculating angles and sides, be sure to carry the exact values through to the final answer. Generally, final answers are rounded to the nearest tenth, unless otherwise specified.

角や辺を計算するときは、最後の答えまで厳密な値を持ち越すよう気をつけること。とくに断らない限り、最後の答えはふつう小数第一位まで丸める。

Law of Sines 正弦定理

Given a triangle with angles and opposite sides labeled as in Figure 6, the ratio of the measurement of an angle to the length of its opposite side will be equal to the other two ratios of angle measure to opposite side. All proportions will be equal. The Law of Sines is based on proportions and is presented symbolically two ways.

図6のように角と対辺に名前を付けた三角形が与えられたとき、一つの角の大きさとその対辺の長さの比は、ほかの二つの角の大きさと対辺の比に等しい。どの比も等しくなる。正弦定理は比にもとづき、記号で二通りに示される。

\frac{\sin α}{a} = \frac{\sin β}{b} = \frac{\sin γ}{c}

\frac{a}{\sin α} = \frac{b}{\sin β} = \frac{c}{\sin γ}

To solve an oblique triangle, use any pair of applicable ratios.

一般の三角形の辺や角を求めるには、当てはまるどの比の組を使ってもよい。

Example 1例1

Solving for Two Unknown Sides and Angle of an AAS TriangleAAS の三角形の未知の二辺と角を求める

Solve the triangle shown in Figure 7 to the nearest tenth.

図7に示した三角形を、小数第一位まで解け。

An oblique triangle with standard labels. Angle alpha is 50 degrees, angle gamma is 30 degrees, and side a is of length 10. Side b is the horizontal base.

Solution 解答

The three angles must add up to 180 degrees. From this, we can determine that

三つの角の和は180度でなければならない。ここから次が決まる。

\begin{array}{l} \begin{array}{l} β = 180° - 50° - 30° \end{array} \\ = 100° \end{array}

To find an unknown side, we need to know the corresponding angle and a known ratio. We know that angle $α = 50°$ and its corresponding side $a = 10.$ We can use the following proportion from the Law of Sines to find the length of $c .$

未知の辺を求めるには、対応する角と、分かっている比が要る。角 $α = 50°$ とその対辺 $a = 10.$ が分かっている。正弦定理から次の比を使えば、 $c$ の長さが求められる。

\begin{array}{l} \frac{\sin (50°)}{10} = \frac{\sin (30°)}{c} \\ c \frac{\sin (50°)}{10} = \sin (30°) & Multiply both sides by c . \\ c = \sin (30 °) \frac{10}{\sin (50°)} & Multiply by the reciprocal to isolate c . \\ c \approx 6.5 \end{array}

Similarly, to solve for $b,$ we set up another proportion.

同じように $b$ について解くために、別の比を立てる。

\begin{array}{l} \begin{array}{l} \frac{\sin (50°)}{10} = \frac{\sin (100°)}{b} \end{array} \\ b \sin (50°) = 10 \sin (100°) & Multiply both sides by b . \\ b = \frac{10 \sin (100°)}{\sin (50°)} \begin{matrix}  \end{matrix} & Multiply by the reciprocal to isolate b . \\ b \approx 12.9 \end{array}

Therefore, the complete set of angles and sides is

したがって、角と辺の完全な組は次のとおりである。

\begin{array}{l} \begin{array}{l} α = 50° a = 10 \end{array} \\ β = 100° b \approx 12.9 \\ γ = 30° c \approx 6.5 \end{array}

Try It #1やってみよう1

Solve the triangle shown in Figure 8 to the nearest tenth.

図8に示した三角形を、小数第一位まで解け。

An oblique triangle with standard labels. Angle alpha is 98 degrees, angle gamma is 43 degrees, and side b is of length 22. Side b is the horizontal base.

Using The Law of Sines to Solve SSA Triangles正弦定理を使って SSA の三角形を解く

We can use the Law of Sines to solve any oblique triangle, but some solutions may not be straightforward. In some cases, more than one triangle may satisfy the given criteria, which we describe as an ambiguous case. Triangles classified as SSA, those in which we know the lengths of two sides and the measurement of the angle opposite one of the given sides, may result in one or two solutions, or even no solution.

正弦定理を使えばどの斜三角形も解けるが、解がまっすぐでないこともある。与えられた条件を満たす三角形が二つ以上ある場合もあり、これをあいまいな場合という。SSA に分けられる三角形、すなわち二辺の長さと、与えられた辺の一方の向かいの角の大きさが分かっているものは、解が一つか二つ、あるいは解が無いこともある。

Possible Outcomes for SSA Triangles 2辺とその間にない角から三角形を求める場合

Oblique triangles in the category SSA may have four different outcomes. Figure 9 illustrates the solutions with the known sides $a$ and $b$ and known angle $α .$

SSA に分けられる斜三角形には、四通りの結果がありうる。図9に、分かっている辺 $a$ と $b$ 、分かっている角 $α$ での解を示す。

Four attempted oblique triangles are in a row, all with standard labels. Side c is the horizontal base. In the first attempted triangle, side a is less than the altitude height. Since side a cannot reach side c, there is no triangle. In the second attempted triangle, side a is equal to the length of the altitude height, so side a forms a right angle with side c. In the third attempted triangle, side a is greater than the altitude height and less than side b, so side a can form either an acute or obtuse angle with side c. In the fourth attempted triangle, side a is greater than or equal to side b, so side a forms an acute angle with side c.

Example 2例2

Solving an Oblique SSA TriangleSSA の一般の三角形の辺や角を求める

Solve the triangle in Figure 10 for the missing side and find the missing angle measures to the nearest tenth.

図10の三角形について欠けている辺を求め、欠けている角の大きさを小数第一位まで求めよ。

An oblique triangle with standard labels where side a is of length 6, side b is of length 8, and angle alpha is 35 degrees.

Solution 解答

Use the Law of Sines to find angle $β$ and angle $γ,$ and then side $c .$ Solving for $β,$ we have the proportion

正弦定理を使って角 $β$ と角 $γ$ 、そして辺 $c$ を求める。 $β$ について解くために、次の比を立てる。

\begin{array}{r} \frac{\sin α}{a} = \frac{\sin β}{b} \\ \frac{\sin (35°)}{6} = \frac{\sin β}{8} \\ \frac{8 \sin (35°)}{6} = \sin β \\ 0.7648 \approx \sin β \\ \sin^{- 1} (0.7648) \approx 49.9° \\ β \approx 49.9° \end{array}

However, in the diagram, angle $β$ appears to be an obtuse angle and may be greater than 90°. How did we get an acute angle, and how do we find the measurement of $β ?$ Let’s investigate further. Dropping a perpendicular from $γ$ and viewing the triangle from a right angle perspective, we have Figure 11. It appears that there may be a second triangle that will fit the given criteria.

だがこの図では、角 $β$ は鈍角に見え、90°より大きいかもしれない。なぜ鋭角が得られたのか。 $β ?$ の大きさはどう求めればよいか。もっと調べてみよう。 $γ$ から垂線を下ろし、直角の見方でこの三角形を見ると、図11のようになる。与えられた条件に合う二つ目の三角形があるように見える。

An oblique triangle built from the previous with standard prime labels. Side a is of length 6, side b is of length 8, and angle alpha prime is 35 degrees. An isosceles triangle is attached, using side a as one of its congruent legs and the angle supplementary to angle beta as one of its congruent base angles. The other congruent angle is called beta prime, and the entire new horizontal base, which extends from the original side c, is called c prime. There is a dotted altitude line from angle gamma prime to side c prime.

The angle supplementary to $β$ is approximately equal to 49.9°, which means that $β = 180° - 49.9° = 130.1° .$ (Remember that the sine function is positive in both the first and second quadrants.) Solving for $γ,$ we have

$β$ の補角はおよそ49.9°に等しい。つまり $β = 180° - 49.9° = 130.1°$ である。(正弦関数は第1象限でも第2象限でも正であることを思い出そう。) $γ$ について解くと、次が得られる。

γ = 180° - 35° - 130.1° \approx 14.9°

We can then use these measurements to solve the other triangle. Since $γ^{'}$ is supplementary to the sum of $α^{'}$ and $β^{'},$ we have

そしてこの値を使ってもう一方の三角形が解ける。 $γ^{'}$ は $α^{'}$ と $β^{'}$ の和の補角なので、次が得られる。

γ^{'} = 180° - 35° - 49.9° \approx 95.1°

Now we need to find $c$ and $c^{'} .$

次に $c$ と $c^{'}$ を求める必要がある。

We have

次が得られる。

\begin{array}{l} \frac{c}{\sin (14.9°)} = \frac{6}{\sin (35°)} \\ c = \frac{6 \sin (14.9°)}{\sin (35°)} \approx 2.7 \end{array}

Finally,

最後に、

\begin{array}{l} \frac{c^{'}}{\sin (95.1°)} = \frac{6}{\sin (35°)} \\ c^{'} = \frac{6 \sin (95.1°)}{\sin (35°)} \approx 10.4 \end{array}

To summarize, there are two triangles with an angle of 35°, an adjacent side of 8, and an opposite side of 6, as shown in Figure 12.

まとめると、図12に示すとおり、35°の角、隣辺8、対辺6を持つ三角形が二つある。

There are two triangles with standard labels. Triangle a is the orginal triangle. It has angles alpha of 35 degrees, beta of 130.1 degrees, and gamma of 14.9 degrees. It has sides a = 6, b = 8, and c is approximately 2.7. Triangle b is the extended triangle. It has angles alpha prime = 35 degrees, angle beta prime = 49.9 degrees, and angle gamma prime = 95.1 degrees. It has side a prime = 6, side b prime = 8, and side c prime is approximately 10.4.

However, we were looking for the values for the triangle with an obtuse angle $β .$ We can see them in the first triangle (a) in Figure 12.

だが求めていたのは鈍角 $β$ を持つ三角形の値である。それは図12の一つ目の三角形(a)に見える。

Try It #2やってみよう2

Given $α = 80°, a = 120,$ and $b = 121,$ find the missing side and angles. If there is more than one possible solution, show both.

$α = 80°, a = 120$ と $b = 121$ が与えられたとき、欠けている辺と角を求めよ。ありうる解が二つ以上あるなら、両方を示せ。

Example 3例3

Solving for the Unknown Sides and Angles of a SSA TriangleSSA の三角形の未知の辺と角を求める

In the triangle shown in Figure 13, solve for the unknown side and angles. Round your answers to the nearest tenth.

図13に示した三角形について、未知の辺と角を求めよ。答えは小数第一位まで丸めよ。

An oblique triangle with standard labels. Side b is 9, side c is 12, and angle gamma is 85. Angle alpha, angle beta, and side a are unknown.

Solution 解答

In choosing the pair of ratios from the Law of Sines to use, look at the information given. In this case, we know the angle $γ = 85°,$ and its corresponding side $c = 12,$ and we know side $b = 9.$ We will use this proportion to solve for $β .$

正弦定理のどの比の組を使うかを選ぶには、与えられた情報を見る。この場合、角 $γ = 85°$ とその対辺 $c = 12$ が分かっており、辺 $b = 9.$ も分かっている。この比を使って $β$ について解く。

\begin{array}{l} \frac{\sin (85°)}{12} = \frac{\sin β}{9} \begin{matrix}  \end{matrix} & Isolate the unknown . \\ \frac{9 \sin (85°)}{12} = \sin β \end{array}

To find $β,$ apply the inverse sine function. The inverse sine will produce a single result, but keep in mind that there may be two values for $β .$ It is important to verify the result, as there may be two viable solutions, only one solution (the usual case), or no solutions.

$β$ を求めるには逆正弦関数を当てはめる。逆正弦は結果を一つだけ返すが、 $β$ の値が二つあるかもしれないことを心にとめておこう。成り立つ解が二つあることも、解が一つだけ(ふつうの場合)のことも、解が無いこともあるので、結果を確かめることが大切である。

\begin{array}{l} β = \sin^{- 1} (\frac{9 \sin (85°)}{12}) \\ β \approx \sin^{- 1} (0.7471) \\ β \approx 48.3° \end{array}

In this case, if we subtract $β$ from 180°, we find that there may be a second possible solution. Thus, $β = 180° - 48.3° \approx 131.7° .$ To check the solution, subtract both angles, 131.7° and 85°, from 180°. This gives

この場合、180°から $β$ を引くと、二つ目のありうる解があるかもしれないと分かる。よって $β = 180° - 48.3° \approx 131.7°$ である。この解を確かめるために、131.7°と85°の両方の角を180°から引く。すると次が得られる。

α = 180° - 85° - 131.7° \approx - 36.7°,

which is impossible, and so $β \approx 48.3° .$

これはありえないので、 $β \approx 48.3°$ である。

To find the remaining missing values, we calculate $α = 180° - 85° - 48.3° \approx 46.7° .$ Now, only side $a$ is needed. Use the Law of Sines to solve for $a$ by one of the proportions.

残りの欠けている値を求めるために $α = 180° - 85° - 48.3° \approx 46.7°$ を計算する。これで辺 $a$ だけが残る。正弦定理の比の一つを使って $a$ について解く。

\begin{array}{l} \begin{array}{l} \begin{array}{l} \frac{\sin (85 °)}{12} = \frac{\sin (46.7 °)}{a} \end{array} \end{array} \\ a \frac{\sin (85 °)}{12} = \sin (46.7 °) \\ a = \frac{12 \sin (46.7 °)}{\sin (85 °)} \approx 8.8 \end{array}

The complete set of solutions for the given triangle is

与えられた三角形の解の完全な組は次のとおりである。

\begin{array}{l} \begin{array}{l} α \approx 46.7° a \approx 8.8 \end{array} \\ β \approx 48.3° b = 9 \\ γ = 85° c = 12 \end{array}

Try It #3やってみよう3

Given $α = 80°, a = 100, b = 10,$ find the missing side and angles. If there is more than one possible solution, show both. Round your answers to the nearest tenth.

$α = 80°, a = 100, b = 10$ が与えられたとき、欠けている辺と角を求めよ。ありうる解が二つ以上あるなら、両方を示せ。答えは小数第一位まで丸めよ。

Example 4例4

Finding the Triangles That Meet the Given Criteria与えられた条件を満たす三角形を求める

Find all possible triangles if one side has length 4 opposite an angle of 50°, and a second side has length 10.

一辺の長さが4で50°の角の向かいにあり、もう一辺の長さが10であるとき、ありうる三角形をすべて求めよ。

Solution 解答

Using the given information, we can solve for the angle opposite the side of length 10. See Figure 14.

与えられた情報を使えば、長さ10の辺の向かいの角について解ける。図14を見よ。

\begin{array}{l} \frac{\sin α}{10} = \frac{\sin (50°)}{4} \\ \sin α = \frac{10 \sin (50°)}{4} \\ \sin α \approx 1.915 \end{array}

An incomplete triangle. One side has length 4 opposite a 50 degree angle, and a second side has length 10 opposite angle a. The side of length 4 is too short to reach the side of length 10, so there is no third angle.

We can stop here without finding the value of $α .$ Because the range of the sine function is $[- 1, 1],$ it is impossible for the sine value to be 1.915. In fact, inputting $\sin^{- 1} (1.915)$ in a graphing calculator generates an ERROR DOMAIN. Therefore, no triangles can be drawn with the provided dimensions.

$α$ の値を求めずに、ここで止められる。正弦関数の値域は $[- 1, 1]$ なので、正弦の値が1.915になることはありえない。実際、グラフ電卓に $\sin^{- 1} (1.915)$ と入れると ERROR DOMAIN になる。したがって、与えられた寸法で描ける三角形は無い。

Try It #4やってみよう4

Determine the number of triangles possible given $a = 31,$ $b = 26,$ $β = 48° .$

$a = 31$ 、 $b = 26$ 、 $β = 48°$ が与えられたとき、ありうる三角形の数を決めよ。

Finding the Area of an Oblique Triangle Using the Sine Function正弦関数を使って一般の三角形の面積を求める

Now that we can solve a triangle for missing values, we can use some of those values and the sine function to find the area of an oblique triangle. Recall that the area formula for a triangle is given as $Area = \frac{1}{2} b h,$ where $b$ is base and $h$ is height. For oblique triangles, we must find $h$ before we can use the area formula. Observing the two triangles in Figure 15, one acute and one obtuse, we can drop a perpendicular to represent the height and then apply the trigonometric property $\sin α = \frac{opposite}{hypotenuse}$ to write an equation for area in oblique triangles. In the acute triangle, we have $\sin α = \frac{h}{c}$ or $c \sin α = h .$ However, in the obtuse triangle, we drop the perpendicular outside the triangle and extend the base $b$ to form a right triangle. The angle used in calculation is $α^{'},$ or $180 - α .$

三角形の欠けている値が求められるようになったので、その値のいくつかと正弦関数を使って一般の三角形の面積を求められる。三角形の面積の公式が $面積 = \frac{1}{2} b h$ で与えられることを思い出そう。ここで $b$ は底辺、 $h$ は高さである。斜三角形では、面積の公式を使う前に $h$ を求めなければならない。図15の二つの三角形——一つは鋭角三角形、もう一つは鈍角三角形——を見ると、高さを表す垂線を下ろし、そして三角の性質 $\sin α = \frac{対辺}{斜辺}$ を当てはめて、一般の三角形の面積の式が書ける。鋭角三角形では $\sin α = \frac{h}{c}$ 、すなわち $c \sin α = h$ である。だが鈍角三角形では、垂線を三角形の外に下ろし、底辺 $b$ を延ばして直角三角形を作る。計算に使う角は $α^{'}$ 、すなわち $180 - α$ である。

Two oblique triangles with standard labels. Both have a dotted altitude line h extended from angle beta to the horizontal base side b. In the first, which is an acute triangle, the altitude is within the triangle. In the second, which is an obtuse triangle, the altitude h is outside of the triangle.

Thus,

したがって、

Area = \frac{1}{2} (base) (height) = \frac{1}{2} b (c \sin α)

Similarly,

同じように、

Area = \frac{1}{2} a (b \sin γ) = \frac{1}{2} a (c \sin β)

Area of an Oblique Triangle 一般の三角形の面積

The formula for the area of an oblique triangle is given by

一般の三角形の面積の公式は次で与えられる。

\begin{array}{l} Area = \frac{1}{2} b c \sin α \\ = \frac{1}{2} a c \sin β \\ = \frac{1}{2} a b \sin γ \end{array}

This is equivalent to one-half of the product of two sides and the sine of their included angle.

これは、二辺の積とそのあいだの角の正弦の積の半分に等しい。

Example 5例5

Finding the Area of an Oblique Triangle一般の三角形の面積を求める

Find the area of a triangle with sides $a = 90, b = 52,$ and angle $γ = 102° .$ Round the area to the nearest integer.

辺が $a = 90, b = 52$ 、角が $γ = 102°$ の三角形の面積を求めよ。面積は整数に丸めよ。

Solution 解答

Using the formula, we have

この公式を使うと、次が得られる。

\begin{array}{l} Area = \frac{1}{2} a b \sin γ \\ Area = \frac{1}{2} (90) (52) \sin (102°) \\ Area \approx 2289 square units \end{array}

Try It #5やってみよう5

Find the area of the triangle given $β = 42°,$ $a = 7.2 ft,$ $c = 3.4 ft .$ Round the area to the nearest tenth.

$β = 42°$ 、 $a = 7.2 ft$ 、 $c = 3.4 ft$ が与えられた三角形の面積を求めよ。面積は小数第一位まで丸めよ。

Solving Applied Problems Using the Law of Sines正弦定理を使って応用問題を解く

The more we study trigonometric applications, the more we discover that the applications are countless. Some are flat, diagram-type situations, but many applications in calculus, engineering, and physics involve three dimensions and motion.

三角法の応用を学べば学ぶほど、その応用が数え切れないと分かってくる。平面の図のような状況もあるが、微積分・工学・物理の多くの応用は三次元と運動を伴う。

Example 6例6

Finding an Altitude高度を求める

Find the altitude of the aircraft in the problem introduced at the beginning of this section, shown in Figure 16. Round the altitude to the nearest tenth of a mile.

図16に示す、この節の初めに出した問題の航空機の高度を求めよ。高度はマイルの小数第一位まで丸めよ。

Solution 解答

To find the elevation of the aircraft, we first find the distance from one station to the aircraft, such as the side $a,$ and then use right triangle relationships to find the height of the aircraft, $h .$

航空機の高度を求めるために、まず一つの局から航空機までの距離、すなわち辺 $a$ を求め、そして直角三角形の関係を使って航空機の高さ $h$ を求める。

Because the angles in the triangle add up to 180 degrees, the unknown angle must be 180°−15°−35°=130°. This angle is opposite the side of length 20, allowing us to set up a Law of Sines relationship.

三角形の角の和は180度なので、未知の角は 180°−15°−35°=130° でなければならない。この角は長さ20の辺の向かいにあるので、正弦定理の関係が立てられる。

\begin{array}{l} \begin{array}{l} \begin{array}{l}  \end{array} \\ \frac{\sin (130°)}{20} = \frac{\sin (35°)}{a} \end{array} \\ a \sin (130°) = 20 \sin (35°) \\ a = \frac{20 \sin (35°)}{\sin (130°)} \\ a \approx 14.98 \end{array}

The distance from one station to the aircraft is about 14.98 miles.

一つの局から航空機までの距離はおよそ14.98マイルである。

Now that we know $a,$ we can use right triangle relationships to solve for $h .$

$a$ が分かったので、直角三角形の関係を使って $h$ について解ける。

\begin{array}{l} \sin (15°) = \frac{opposite}{hypotenuse} \\ \sin (15°) = \frac{h}{a} \\ \sin (15°) = \frac{h}{14.98} \\ h = 14.98 \sin (15°) \\ h \approx 3.88 \end{array}

The aircraft is at an altitude of approximately 3.9 miles.

この航空機の高度はおよそ3.9マイルである。

Try It #6やってみよう6

The diagram shown in Figure 17 represents the height of a blimp flying over a football stadium. Find the height of the blimp if the angle of elevation at the southern end zone, point A, is 70°, the angle of elevation from the northern end zone, point $B,$ is 62°, and the distance between the viewing points of the two end zones is 145 yards.

図17に示した図は、競技場の上を飛ぶ飛行船の高さを表している。南のエンドゾーンの点Aでの仰角が70°、北のエンドゾーンの点 $B$ からの仰角が62°、二つのエンドゾーンの観測点のあいだの距離が145ヤードのとき、この飛行船の高さを求めよ。

An oblique triangle formed from three vertices A, B, and C. Verticies A and B are points on the ground, and vertex C is the blimp in the air between them. The distance between A and B is 145 yards. The angle at vertex A is 70 degrees, and the angle at vertex B is 62 degrees.

Media 教材

Access these online resources for additional instruction and practice with trigonometric applications.

三角法の応用についての説明と練習をさらに求めるなら、次のオンライン教材にあたるとよい。

• Law of Sines: The Basics• Law of Sines: The Ambiguous Case

・正弦定理: 基本・正弦定理: あいまいな場合

Verbal言葉で答える問題

1. Describe the altitude of a triangle.

1. 三角形の高さを述べよ。

2. Compare right triangles and oblique triangles.

2. 直角三角形と斜三角形を比べよ。

3. When can you use the Law of Sines to find a missing angle?

3. 正弦定理はどんなときに欠けている角を求めるのに使えるか。

4. In the Law of Sines, what is the relationship between the angle in the numerator and the side in the denominator?

4. 正弦定理で、分子の角と分母の辺のあいだにはどんな関係があるか。

5. What type of triangle results in an ambiguous case?

5. どんな種類の三角形があいまいな場合になるか。

Algebraic代数の問題

For the following exercises, assume $α$ is opposite side $a, β$ is opposite side $b,$ and $γ$ is opposite side $c .$ Solve each triangle, if possible. Round each answer to the nearest tenth.

次の各問では、 $α$ が辺 $a, β$ の向かい、 $b$ が辺 $γ$ の向かい、 $c$ の向かいだとする。できるなら各三角形を解け。各答えを小数第一位まで丸めよ。

6. $α = 43°, γ = 69°, a = 20$

7. $α = 35°, γ = 73°, c = 20$

8. $α = 60°,$ $β = 60°,$ $γ = 60°$

9. $a = 4,$ $α = 60°,$ $β = 100°$

10. $b = 10,$ $β = 95°, γ = 30°$

For the following exercises, use the Law of Sines to solve for the missing side for each oblique triangle. Round each answer to the nearest hundredth. Assume that angle $A$ is opposite side $a,$ angle $B$ is opposite side $b,$ and angle $C$ is opposite side $c .$

次の各問について、正弦定理を使って各斜三角形の欠けている辺を求めよ。各答えを小数第二位まで丸めよ。角 $A$ が辺 $a$ の向かい、角 $B$ が辺 $b$ の向かい、角 $C$ が辺 $c$ の向かいだとする。

11. Find side $b$ when $A = 37°,$ $B = 49°,$ $c = 5.$

11. $A = 37°$ 、 $B = 49°$ 、 $c = 5.$ のとき、辺 $b$ を求めよ。

12. Find side $a$ when $A = 132°, C = 23°, b = 10.$

12. $A = 132°, C = 23°, b = 10.$ のとき、辺 $a$ を求めよ。

13. Find side $c$ when $B = 37°, C = 21°,$ $b = 23.$

13. $B = 37°, C = 21°$ 、 $b = 23.$ のとき、辺 $c$ を求めよ。

For the following exercises, assume $α$ is opposite side $a, β$ is opposite side $b,$ and $γ$ is opposite side $c .$ Determine whether there is no triangle, one triangle, or two triangles. Then solve each triangle, if possible. Round each answer to the nearest tenth.

次の各問では、 $α$ が辺 $a, β$ の向かい、 $b$ が辺 $γ$ の向かい、 $c$ の向かいだとする。三角形が無いか、一つか、二つかを判定せよ。そしてできるなら各三角形を解け。各答えを小数第一位まで丸めよ。

14. $α = 119°, a = 14, b = 26$

15. $γ = 113°, b = 10, c = 32$

16. $b = 3.5,$ $c = 5.3,$ $γ = 80°$

17. $a = 12,$ $c = 17,$ $α = 35°$

18. $a = 20.5,$ $b = 35.0,$ $β = 25°$

19. $a = 7,$ $c = 9,$ $α = 43°$

20. $a = 7, b = 3, β = 24°$

21. $b = 13, c = 5, γ = 10°$

22. $a = 2.3, c = 1.8, γ = 28°$

23. $β = 119°, b = 8.2, a = 11.3$

For the following exercises, use the Law of Sines to solve, if possible, the missing side or angle for each triangle or triangles in the ambiguous case. Round each answer to the nearest tenth.

次の各問について、正弦定理を使い、できるなら各三角形、あるいはあいまいな場合の三角形の欠けている辺や角を求めよ。各答えを小数第一位まで丸めよ。

24. Find angle $A$ when $a = 24, b = 5, B = 22°.$

24. $a = 24, b = 5, B = 22°.$ のとき、角 $A$ を求めよ。

25. Find angle $A$ when $a = 13, b = 6, B = 20°.$

25. $a = 13, b = 6, B = 20°.$ のとき、角 $A$ を求めよ。

26. Find angle $B$ when $A = 12°, a = 2, b = 9.$

26. $A = 12°, a = 2, b = 9.$ のとき、角 $B$ を求めよ。

For the following exercises, find the area of the triangle with the given measurements. Round each answer to the nearest tenth.

次の各問について、与えられた寸法の三角形の面積を求めよ。各答えを小数第一位まで丸めよ。

27. $a = 5, c = 6, β = 35°$

28. $b = 11, c = 8, α = 28°$

29. $a = 32, b = 24, γ = 75°$

30. $a = 7.2, b = 4.5, γ = 43°$

Graphicalグラフの問題

For the following exercises, find the length of side $x .$ Round to the nearest tenth.

次の各問について、辺 $x$ の長さを求めよ。小数第一位まで丸めよ。

31.

A triangle with an angle of 50 degrees and opposite side of length 10. Another angle is 70 degrees with side opposite of length x.

32.

A triangle with one angle = 120 degrees. Another angle is 25 degrees with side opposite = x. The side adjacent to the 25 and 120 degree angles is of length 6.

33.

A triangle. One angle is 45 degrees with side opposite = x. Another angle is 75 degrees. The side adjacent to the 45 and 75 degree angles = 15.

34.

A triangle. One angle is 40 degrees with opposite side = x. Another angle is 110 degrees with side opposite = 18.

35.

A triangle. One angle is 50 degrees with opposite = x. Another angle is 42 degrees with opposite side = 14.

36.

A triangle. One angle is 111 degrees with opposite side = x. Another angle is 22 degrees. The side adjacent to the 111 and 22 degree angles = 8.6.

For the following exercises, find the measure of angle $x,$ if possible. Round to the nearest tenth.

次の各問について、できるなら角 $x$ の大きさを求めよ。小数第一位まで丸めよ。

37.

A triangle. One angles is 98 degrees with opposite side = 10. Another angle is x degrees with opposite side = 5.

38.

A triangle. One angle is 37 degrees with opposite side = 11. Another angle is x degrees with opposite side = 8.

39.

A triangle. One angle is 22 degrees with side opposite = 5. Another angle is x degrees with opposite side = 13.

40.

A triangle. One angle is 59 degrees with opposite side = 5.7. Another angle is x degrees with opposite side = 5.3.

41. Notice that $x$ is an obtuse angle.

41. $x$ が鈍角であることに注意せよ。

A triangle. One angle is 55 degrees with side opposite = 21. Another angle is x degrees with opposite side = 24.

42.

A triangle. One angle is 65 degrees with opposite side = 10. Another angle is x degrees with opposite side = 12.

For the following exercise, solve the triangle. Round each answer to the nearest tenth.

次の問について、この三角形を解け。各答えを小数第一位まで丸めよ。

43.

A triangle. One angle is 93 degrees with opposite side = 32.6. Another side is 24.1.

44. For the following exercises, find the area of each triangle. Round each answer to the nearest tenth.

44. 次の各問について、各三角形の面積を求めよ。各答えを小数第一位まで丸めよ。

A triangle. One angle is 30 degrees. The two sides adjacent to that angle are 10 and 16.

45.

A triangle. One angle is 25 degrees. The two sides adjacent to that angle are 18 and 15

46.

A triangle. One angle is 51 degrees with opposite side = 3.5. The other two sides are 4.5 and 2.9.

47.

A triangle. One angle is 58 degrees with opposite side unknown. Another angle is 51 degrees with opposite side = 9. The side adjacent to the two given angles is 11.

48.

A triangle. One angle is 40 degrees with opposite side = 18. One of the other sides is 25.

49.

A triangle. One angle is 115 degrees with opposite side = 50. Another angle is 30 degrees with opposite side = 30.

Extensions発展問題

50. Find the radius of the circle in Figure 18. Round to the nearest tenth.

50. 図18の円の半径を求めよ。小数第一位まで丸めよ。

A triangle inscribed in a circle. Two of the legs are radii. The central angle formed by the radii is 145 degrees, and the opposite side is 3.

51. Find the diameter of the circle in Figure 19. Round to the nearest tenth.

51. 図19の円の直径を求めよ。小数第一位まで丸めよ。

A triangle inscribed in a circle. Two of the legs are radii. The central angle formed by the radii is 110 degrees, and the opposite side is 8.3.

52. Find $m ∠ A D C$ in Figure 20. Round to the nearest tenth.

52. 図20の $m ∠ A D C$ を求めよ。小数第一位まで丸めよ。

A triangle inside a triangle. The outer triangle is formed by vertices A, B, and D. Side B D is the base. The inner triangle shares vertices A and B. The last vertex C is located on the base side of the outer triangle between vertices B and D. Angle B is 60 degrees, side A D is 10, and side A C is 9.

53. Find $A D$ in Figure 21. Round to the nearest tenth.

53. 図21の $A D$ を求めよ。小数第一位まで丸めよ。

54. Solve both triangles in Figure 22. Round each answer to the nearest tenth.

54. 図22の二つの三角形を解け。各答えを小数第一位まで丸めよ。

Two triangles formed by intersecting lines A D and B C. They intersect at point E. The first triangle is formed from vertices A, B, and E while the second triangle is formed from vertices C, E, and D. Angle A is 48 degrees, side A B is 4.2, angle D is 48 degrees, and side C D is 2. Angle A E B is 46 degrees.

55. Find $A B$ in the parallelogram shown in Figure 23.

55. 図23に示した平行四辺形の $A B$ を求めよ。

A parallelogram with vertices A, B, C, and D. There is a diagonal from vertex B to vertex C. Angle A is 130 degrees, angle D is 130 degrees, side B D is 10, and the diagonal B C is 12.

56. Solve the triangle in Figure 24. (Hint: Draw a perpendicular from $H$ to $J K).$ Round each answer to the nearest tenth.

56. 図24の三角形を解け。(ヒント: $H$ から $J K).$ へ垂線を引け。)各答えを小数第一位まで丸めよ。

A triangle with vertices J, K, and H. Side J K is the horizontal base and is 10. Side JH is 7. Angle J is 20 degrees.

57. Solve the triangle in Figure 25. (Hint: Draw a perpendicular from $N$ to $L M).$ Round each answer to the nearest tenth.

57. 図25の三角形を解け。(ヒント: $N$ から $L M).$ へ垂線を引け。)各答えを小数第一位まで丸めよ。

A triangle with vertices M, N, and L. Side M N is the horizontal base and is 4.6. Angle M is 74 degrees, and side M L is 5.

58. In Figure 26, $A B C D$ is not a parallelogram. $∠ m$ is obtuse. Solve both triangles. Round each answer to the nearest tenth.

58. 図26で、 $A B C D$ は平行四辺形ではない。 $∠ m$ は鈍角である。二つの三角形を解け。各答えを小数第一位まで丸めよ。

A quadrilateral with vertices A, B, C, and D. There is a diagonal from vertex B to vertex D of length 45. Side A B is x, side B C is y, side C D is 40, and side D A is 29. Angle A is m degrees, angle C is 65 degrees, angle A B D is 35 degrees, angle D B C is n degrees, angle B D C is k degrees, and angle A D B is h degrees.

Real-World Applications応用問題

59. A pole leans away from the sun at an angle of $7°$ to the vertical, as shown in Figure 27. When the elevation of the sun is $55°,$ the pole casts a shadow 42 feet long on the level ground. How long is the pole? Round the answer to the nearest tenth.

59. 図27に示すとおり、ある柱が鉛直から $7°$ の角で太陽と反対の向きに傾いている。太陽の高度が $55°$ のとき、この柱は平らな地面に42フィートの影を落とす。この柱の長さはいくらか。答えを小数第一位まで丸めよ。

A triangle within a triangle. The outer triangle is formed by vertices A, B, and S (the sun). Side A B is the horizontal base, the ground, and is 42 feet. Angle A is 55 degrees. The inner triangle is formed by vertices A, B, and C. Side B C is the pole. Vertex C is located on side A S of the outer triangle between vertices A and S. Angle C B S is 7 degrees.

60. To determine how far a boat is from shore, two radar stations 500 feet apart find the angles out to the boat, as shown in Figure 28. Determine the distance of the boat from station $A$ and the distance of the boat from shore. Round your answers to the nearest whole foot.

60. 船が岸からどれだけ離れているかを決めるため、図28に示すとおり、500フィート離れた二つのレーダー局が船への角を求める。局 $A$ から船までの距離と、船から岸までの距離を決めよ。答えはフィートの整数に丸めよ。

A triangle formed by the two radar stations A and B and the boat. Side A B is the horizontal base. Angle A is 70 degrees and angle B is 60 degrees.

61. Figure 29 shows a satellite orbiting Earth. The satellite passes directly over two tracking stations $A$ and $B,$ which are 69 miles apart. When the satellite is on one side of the two stations, the angles of elevation at $A$ and $B$ are measured to be $83.9°$ and $86.2°,$ respectively. How far is the satellite from station $A$ and how high is the satellite above the ground? Round answers to the nearest whole mile.

61. 図29は地球を回る人工衛星を示している。この人工衛星は、69マイル離れた二つの追跡局 $A$ と $B$ の真上を通る。人工衛星が二つの局の片側にあるとき、 $A$ と $B$ での仰角はそれぞれ $83.9°$ と $86.2°$ と測られる。人工衛星は局 $A$ からどれだけ離れているか。また、地上どれだけの高さにあるか。答えはマイルの整数に丸めよ。

A triangle formed by two ground tracking stations A and B and the satellite. Side A B is the horizontal base of the triangle. Angle A is 83.9 degrees, and the supplementary angle to angle B is 86.2 degrees.

62. A communications tower is located at the top of a steep hill, as shown in Figure 30. The angle of inclination of the hill is $67° .$ A guy wire is to be attached to the top of the tower and to the ground, 165 meters downhill from the base of the tower. The angle formed by the guy wire and the hill is $16°.$ Find the length of the cable required for the guy wire to the nearest whole meter.

62. 図30に示すとおり、通信塔が急な丘の頂上にある。この丘の傾きの角は $67°$ である。塔の頂上と、塔の根元から丘を165メートル下った地面とに支線を張る。支線と丘がなす角は $16°.$ である。この支線に要る索の長さを、メートルの整数に丸めて求めよ。

A triangle formed by the bottom of the hill, the base of the tower at the top of the hill, and the top of the tower. The side between the bottom of the hill and the top of the tower is wire. The length of the side bertween the bottom of the hill and the bottom of the tower is 165 meters. The angle formed by the wire side and the bottom of the hill is 16 degrees. The angle between the hill and the horizontal ground is 67 degrees.

63. The roof of a house is at a $20°$ angle. An 8-foot solar panel is to be mounted on the roof and should be angled $38°$ relative to the horizontal for optimal results. (See Figure 31). How long does the vertical support holding up the back of the panel need to be? Round to the nearest tenth.

63. ある家の屋根は $20°$ の角である。8フィートの太陽電池板を屋根に取り付け、最も良い結果を得るために水平から $38°$ の角にする。(図31を見よ。)板の後ろを支える縦の支柱はどれだけの長さが要るか。小数第一位まで丸めよ。

A triangle whose sides are the solar panel, the roof which goes past the solar panel, and the vertical support for the panel. The solar panel side is 8 feet long. There are horizontal dotted lines at the bottom of the solar panel and the bottom of the roof. The angle between the solar panel and the horizontal is 38 degrees. The angle between the roof and the horizontal is 20 degrees.

64. Similar to an angle of elevation, an angle of depression is the acute angle formed by a horizontal line and an observer’s line of sight to an object below the horizontal. A pilot is flying over a straight highway. He determines the angles of depression to two mileposts, 6.6 km apart, to be $37°$ and $44°,$ as shown in Figure 32. Find the distance of the plane from point $A$ to the nearest tenth of a kilometer.

64. 仰角と同じように、俯角とは、水平線と、水平より下の物への観測者の視線がなす鋭角である。ある操縦士がまっすぐな幹線道路の上を飛んでいる。図32に示すとおり、6.6km離れた二つのマイル標識への俯角が $37°$ と $44°$ だと定める。点 $A$ から飛行機までの距離を、キロメートルの小数第一位まで求めよ。

A triangle formed by points A and B on the ground and a plane in the air between them. Side A B is the horizontal ground. There is a horizontal dotted line parallel to the ground going through the plane. The angle formed by the dotted horizontal, the plane, and point A is 37 degrees. The angle between the dotted horizontal, the plane, and point B is 44 degrees.

65. A pilot is flying over a straight highway. He determines the angles of depression to two mileposts, 4.3 km apart, to be 32° and 56°, as shown in Figure 33. Find the distance of the plane from point $A$ to the nearest tenth of a kilometer.

65. ある操縦士がまっすぐな幹線道路の上を飛んでいる。図33に示すとおり、4.3km離れた二つのマイル標識への俯角が32°と56°だと定める。点 $A$ から飛行機までの距離を、キロメートルの小数第一位まで求めよ。

A triangle formed between the plane and two points on the ground, A and B. Side A B is the horizontal base. The plane is above and to the left of both A and B. Point B is to the right of point A. There is a dotted horizontal line going through the plane parallel to the ground. The angle formed between point B, the plane, and the dotted horizontal line is 32 degrees. The angle formed between point A, the plane, and the dotted horizontal line is 56 degrees.

66. In order to estimate the height of a building, two students stand at a certain distance from the building at street level. From this point, they find the angle of elevation from the street to the top of the building to be 39°. They then move 300 feet closer to the building and find the angle of elevation to be 50°. Assuming that the street is level, estimate the height of the building to the nearest foot.

66. ある建物の高さを見積もるために、二人の学生が路面から建物までのある距離のところに立つ。この点から、路面から建物の頂上への仰角が39°だと分かる。そして建物へ300フィート近づくと、仰角が50°だと分かる。路面が平らだとして、この建物の高さをフィートの整数に丸めて見積もれ。

67. In order to estimate the height of a building, two students stand at a certain distance from the building at street level. From this point, they find the angle of elevation from the street to the top of the building to be 35°. They then move 250 feet closer to the building and find the angle of elevation to be 53°. Assuming that the street is level, estimate the height of the building to the nearest foot.

67. ある建物の高さを見積もるために、二人の学生が路面から建物までのある距離のところに立つ。この点から、路面から建物の頂上への仰角が35°だと分かる。そして建物へ250フィート近づくと、仰角が53°だと分かる。路面が平らだとして、この建物の高さをフィートの整数に丸めて見積もれ。

68. Points $A$ and $B$ are on opposite sides of a lake. Point $C$ is 97 meters from $A .$ The measure of angle $B A C$ is determined to be 101°, and the measure of angle $A C B$ is determined to be 53°. What is the distance from $A$ to $B,$ rounded to the nearest whole meter?

68. 点 $A$ と $B$ は湖の反対側にある。点 $C$ は $A$ から97メートルである。角 $B A C$ の大きさは101°、角 $A C B$ の大きさは53°と決められた。 $A$ から $B$ までの距離は、メートルの整数に丸めていくらか。

69. A man and a woman standing $3 \frac{1}{2}$ miles apart spot a hot air balloon at the same time. If the angle of elevation from the man to the balloon is 27°, and the angle of elevation from the woman to the balloon is 41°, find the altitude of the balloon to the nearest foot.

69. $3 \frac{1}{2}$ マイル離れて立っている男性と女性が、同時に熱気球を見つける。男性から気球への仰角が27°、女性から気球への仰角が41°のとき、この気球の高度をフィートの整数に丸めて求めよ。

70. Two search teams spot a stranded climber on a mountain. The first search team is 0.5 miles from the second search team, and both teams are at an altitude of 1 mile. The angle of elevation from the first search team to the stranded climber is 15°. The angle of elevation from the second search team to the climber is 22°. What is the altitude of the climber? Round to the nearest tenth of a mile.

70. 二つの捜索隊が山で立ち往生した登山者を見つける。一つ目の捜索隊は二つ目の捜索隊から0.5マイル離れており、どちらの隊も高度1マイルにいる。一つ目の捜索隊から立ち往生した登山者への仰角は15°である。二つ目の捜索隊から登山者への仰角は22°である。この登山者の高度はいくらか。マイルの小数第一位まで丸めよ。

71. A street light is mounted on a pole. A 6-foot-tall man is standing on the street a short distance from the pole, casting a shadow. The angle of elevation from the tip of the man’s shadow to the top of his head of 28°. A 6-foot-tall woman is standing on the same street on the opposite side of the pole from the man. The angle of elevation from the tip of her shadow to the top of her head is 28°. If the man and woman are 20 feet apart, how far is the street light from the tip of the shadow of each person? Round the distance to the nearest tenth of a foot.

71. 街灯が柱に取り付けられている。身長6フィートの男性が、柱から少し離れた通りに立って影を落としている。男性の影の先から頭のてっぺんへの仰角は28°である。身長6フィートの女性が、同じ通りの、男性から見て柱の反対側に立っている。彼女の影の先から頭のてっぺんへの仰角は28°である。男性と女性が20フィート離れているとき、街灯は各人の影の先からどれだけ離れているか。距離をフィートの小数第一位まで丸めよ。

72. Three cities, $A, B,$ and $C,$ are located so that city $A$ is due east of city $B .$ If city $C$ is located 35° west of north from city $B$ and is 100 miles from city $A$ and 70 miles from city $B,$ how far is city $A$ from city $B ?$ Round the distance to the nearest tenth of a mile.

72. 三つの都市 $A, B$ と $C$ は、都市 $A$ が都市 $B$ の真東になるように位置している。都市 $C$ が都市 $B$ から北の西35°の向きにあり、都市 $A$ から100マイル、都市 $B$ から70マイルのとき、都市 $A$ は都市 $B ?$ からどれだけ離れているか。距離をマイルの小数第一位まで丸めよ。

73. Two streets meet at an 80° angle. At the corner, a park is being built in the shape of a triangle. Find the area of the park if, along one road, the park measures 180 feet, and along the other road, the park measures 215 feet.

73. 二本の通りが80°の角で交わっている。その角に三角形の形の公園が造られている。一方の道に沿って公園が180フィート、もう一方の道に沿って215フィートのとき、この公園の面積を求めよ。

74. Brian’s house is on a corner lot. Find the area of the front yard if the edges measure 40 and 56 feet, as shown in Figure 34.

74. Brian の家は角地にある。図34に示すとおり、縁が40フィートと56フィートのとき、前庭の面積を求めよ。

A triangle with angle 135 degrees. The sides adjacent to that angle are 56 feet and 40 feet. The other side is the house, length unknown.

75. The Bermuda triangle is a region of the Atlantic Ocean that connects Bermuda, Florida, and Puerto Rico. Find the area of the Bermuda triangle if the distance from Florida to Bermuda is 1030 miles, the distance from Puerto Rico to Bermuda is 980 miles, and the angle created by the two distances is 62°.

75. バミューダ・トライアングルは、バミューダ・フロリダ・プエルトリコを結ぶ大西洋の海域である。フロリダからバミューダまでの距離が1030マイル、プエルトリコからバミューダまでの距離が980マイル、この二つの距離がなす角が62°のとき、バミューダ・トライアングルの面積を求めよ。

76. A yield sign measures 30 inches on all three sides. What is the area of the sign?

76. ある譲れの標識は三辺とも30インチである。この標識の面積はいくらか。

77. Naomi bought a dining table whose top is in the shape of a triangle. Find the area of the table top if two of the sides measure 4 feet and 4.5 feet, and the smaller angles measure 32° and 42°, as shown in Figure 35.

77. Naomi は天板が三角形の食卓を買った。図35に示すとおり、二辺が4フィートと4.5フィート、小さいほうの角が32°と42°のとき、この天板の面積を求めよ。

A triangle. One angle is 32 degrees with opposite side = 4. Another angle is 42 degrees with opposite side = 4.5.