Question

1. From two points 240m apart on a horizontal road running east from a mountain the angles of elevation of its top are, respectively 450 and 300. How high above the road is from the top of the mountain.

Answer 1

Answer on Question#40444 – Math – Geometry

From two points $240\mathrm{m}$ apart on a horizontal road running east from a mountain the angles of elevation of its top are, respectively $45{}^{\circ}$ and $30{}^{\circ}$. How high above the road is from the top of the mountain.

Solution:

A = point at 45 degrees

B = point at 30 degrees

C = point at top of mountain

AB = 240m – length of the horizontal road

Firstly, we need to find $\angle C$.

So

$$\angle C = 180{}^{\circ} - 45{}^{\circ} - 30{}^{\circ} = 105{}^{\circ}$$

Using sine rule:

$$\frac{AB}{\sin(105{}^{\circ})} = \frac{BC}{\sin(45{}^{\circ})}$$

Thus,

$$BC = AB \cdot \frac{\sin 45{}^{\circ}}{\sin 105{}^{\circ}} \quad (1)$$

From the right triangle BCH:

$$\sin 30{}^{\circ} = \frac{h}{BC} \Rightarrow h = BC \sin 30{}^{\circ} \quad (2)$$

(1)in(2):

$$h = AB \cdot \frac{\sin 45{}^{\circ}}{\sin 105{}^{\circ}} \sin 30{}^{\circ} = 240\,\mathrm{m} \cdot \frac{\sin 45{}^{\circ}}{\sin 105{}^{\circ}} \sin 30{}^{\circ} = 87.85\,\mathrm{m}$$

Answer: top of the mountain is $87.85\,\mathrm{m}$ above the road.