Answer to Question #142874 in Trigonometry for Ippi

Question

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, 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.

Answer 1

SOLUTION

Here is the sketch of the scenario;

First, we separate the triangles;

Here, we apply the Sine Rule;

$\frac{a}{sinA}=\frac{b}{sinB}=\frac{c}{sinC}$

In this case we apply a and c. i.e.

$\frac{a}{sinA}=\frac{c}{sinC}$

$\frac{92}{sin(8°)}=\frac{c}{sin(61°)}$

$c = \frac{92×sin(61°)}{sin(8°)}$

$c = 578.16m$

Therefore, $X = 578.16m$

Then, considering the last triangle, which is right-angle triangle, we have;

We consider the SOHCAHTOA concept, where we apply the sine formula (SOH)

$sin(69°)=\frac{H}{578.16}$

$H= 578.16 × sin (69°)$

$H = 539.76m$ (To the nearest tenths)

$H = 539.8m$

Therefore, Height of the cliff is $539.8m$