Question

You and your partner are observing an aircraft from two observation decks, located 5.0 krn apart. From your point of view, the aircraft is at ar angle of elevation of 70°. From your partner's point of view, the angle of elevation is 55º. Determine the altitude of the aircraft, to the nearest tenth ofa kilometre. 7° SS 5.0 km

Answer 1

The altitude of the aircraft observed is 15km to the nearest whole kilometres.

Let the distance of myself from the base of the aircraft vertical height be represented with x while the altitude of the aircraft be y so that by observation we get two right triangles with same opposite sides to the angles formed by my partner and I. We can solve for x and y using tangent of the angles as follows:

tan 70 = y/x {opposite/adjacent}

y = xtan70 {cross multiplication}

tan 55 = y/(5 + x)

y = tan55(5 + x)

Thus;

xtan70 = tan55(5 + x)

xtan70 = 5tan55 + xtan55

xtan70 - xtan55 = 5tan55 {collect like terms}

x(tan70 - tan55) = 5tan55

x = 5tan55/(tan70 - tan55)

x = 5.4124 km

Putting the 44.1230m for x in y = xtan70 we have;

y = 5.4124 km × tan70

y = 14.8704 km.

Therefore, the height of the building approximately to the nearest whole kilometres is 15.