it is a math question particularly trigo

Question

a.   At a particular instant, when an airplane was directly above straight-line road connecting two small towns, the angle of depression of these towns were 10.2˚ and 8.7˚. Find (a) the straight lines distances form the airplane to each of the towns at this instant given that the towns are 8.45 km apart. (b) Determine the height of the plane at this instant.

 

Answer 1

This solution assumes that the place is horizontallhy between the two towns.

Draw and label a diagram:

One town is A, the other tiown is B, the plane is P

Drop a perpendicular from the P r to the gound and label G

Angle depression to A is 8.7°, angle of depressiont to B is 10.2°

Label distance from G to B as d.

Label disgtance from P to G as h

The equations:

sin 10.2 = h / d

d sin 10.2 = h

d = h / (sin 10.2)

sin 8.7 = h / (8.45 - d)

(8.45 - d) sin 8.7 = h

8.45 sin 8.7 - d sin 8.7 = h

8.45 sin 8.7 = h + d sin 8.7

8.45 sin 8.7 - h = d sin 8.7

(8.45 sin 8.7 - h) / sin 8.7 = d

Two equations in d. Set equal to each other and solve for h (a lot of messy arithmetic you can do)

Answer 2

a. Draw a diagram and use the Law of Sines.

Let x and y be the straight line distances from the plane to each town. We have a triangle with sides x, y, and 8.45. The angle opposite 8.45 is 10.2° - 8.7° = 1.5°. The angle opposite y is 8.7° and the angle opposite x is 169.8°.

By the Law of Sines, sin169.8° / x = sin1.5° / 8.45 and sin8.7° / y = sin1.5° / 8.45.

We get x = 57.16 km and y = 48.83 km.

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b. Draw a vertical line segment from the plane to the ground with length h. A right triangle is formed with vertical side h, hypotenuse y , and horizontal side along the ground.

The angle opposite side h is 180° - 169.8° = 10.2°

So, sin10.2° = h / y = h / 48.83. h = 48.83sin10.2° = 8.65 km