At an angle of elevation 610 captain on a ship noticed a person on the top of a cliff waiving towards the ship. To see him better, the captain moved his ship 92m closer to the cliff. The angle of elevation at that
moment was 690. Determine the height of the cliff to the nearest tenth of a meter.

The height of the cliff CD is approximately 539.76 m

Step-by-step explanation:

The given parameters are;

The first angle of elevation with which the captain sees the person on the cliff = 61°

The second angle of elevation with which the captain sees the person on the cliff after moving 92 m closer to the cliff = 69°

The angle made by the adjacent supplementary angle to the second angle of elevation = 180° - 69° = 111°

∴ Whereby, the rays from the first and second angle of elevation and the distance the ship moves closer to the cliff forms an imaginary triangle, we have;

The angle in the imaginary triangle subtended by the distance the ship moves closer to the cliff = 180° - 111° - 61° = 8°

By sine rule, we have;

AB/(sin(a)) = BC/(sin(c))

Which gives;

92/(sin(8°)) = BC/(sin(61°))

BC = (sin(61°)) × 92/(sin(8°)) ≈ 578.165 m

BC ≈ 578.165 m

The height CD = BC × sin(69°)

∴  The height of the cliff CD = 578.165 m × sin(69°) ≈ 539.76 m.

The height of the cliff CD ≈ 539.76 m.

this is answer of 144 cubic meters