From the top of a vertical cliff 80m high, the angles of depression of 2 buoys lying due west of the cliff are 23° and 15° respectively. How far are the buoys apart?

The geometry of the problem can be modeled by a right triangle. The height of the cliff forms one leg, and the distance to a buoy forms the other leg. The angle of depression is opposite the height of the cliff. An appropriate trig relation is ...

Tan = Opposite/Adjacent

Solving for the Adjacent side (the distance to the buoy), we find ...

Adjacent = Opposite/Tan

distance to buoy 1 = (80 m)/tan(23°) ≈ 188.468 m

distance to buoy 2 = (80 m)/tan(15°) ≈ 298.564 m

Then the distance between the buoys is ...

298.564 -188.468 m ≈ 110 m

The buoys are about 110 meters apart.

From the top of a vertical cliff 80m high, the angles of depression of 2 buoys lying due west of the cliff are 23° and 15° respectively. How far are the buoys apart?​

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From the top of a vertical cliff 80m high, the angles of depression of 2 buoys lying due west of the cliff are 23° and 15° respectively. How far are the buoys apart?