Question

A LINE AB along one bank of a stream is 562.0 ft.long and C is a point on the opposite bank. The angle BAC is 53°18' and the angle ABC is 48°36' Find the width of the stream.

Answer 1

Answer on Question #58421 – Math – Trigonometry

Question

A LINE AB along one bank of a stream is 562.0 ft. long and C is a point on the opposite bank. The angle BAC is $53{}^{\circ}18'$ and the angle ABC is $48{}^{\circ}36'$ and the width of the stream.

Solution

First of all, let’s find $z{}^{\circ}$ angle. The sum of interior angles of any triangle is equal to $180{}^{\circ}$. (Notice that we will convert minutes to degrees).

$$\angle BAC = \angle A = 53{}^{\circ}18' = (53{}^{\circ}) \cdot \left(\frac{18}{60}\right){}^{\circ} = 53.3{}^{\circ},$$ $$\angle ABC = \angle B = 48{}^{\circ}36' = (48{}^{\circ}) \cdot \left(\frac{36}{60}\right){}^{\circ} = 48.6{}^{\circ},$$ $$\angle ACB = \angle C = 180 - \angle A - \angle B = 180 - 53.3{}^{\circ} - 48.6 = 78.1{}^{\circ}$$

Next we will use the SINE Law $\left(\frac{\sin\angle C}{AB} = \frac{\sin\angle B}{CA} = \frac{\sin\angle A}{CB}\right)$ to obtain CB:

$$\frac{\sin\angle C}{AB} = \frac{\sin\angle A}{CB},$$ $$CB = AB \cdot \frac{\sin\angle A}{\sin\angle C} = 562.0\,ft \cdot \frac{\sin(53.3{}^{\circ})}{\sin(78.1{}^{\circ})} \approx 460.49\,ft,$$

Let's look at $\Delta CBD$ (right triangle). $CD$ is our goal. We know $\angle B$ angle and hypotenuse $CB$ , so let's use the definition of the sine function:

$$\sin \angle B = \frac {C D}{C B},$$ $$\boldsymbol {C D} = C D \cdot \sin \angle B = 4 6 0. 4 9 \mathrm {f t} \cdot \sin (4 8. 6 {}^ {\circ}) \approx 3 4 5. 4 2 \mathrm {f t}.$$

ANSWER: Width of the stream is equal to 345.42 ft.

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