Question

1. sheryl is surveying a cliff to determine the elevation from the base of a canyon to the top of the cliff. she lays out a line ab that is 225 m in length. she also sites a point c at the base of the cliff. point d is a point directly above point c, at the top of the cliff. she measures ∠cab to be 43°, ∠cba to be 58°, and the angle of elevation from point a to point d to be 29°. solve for the height x of the cliff, to the nearest tenth of a metre. 5t

Explanation:

Step1: Find the length of \( AC \) using the Law of Sines in \( \triangle ABC \)

In \( \triangle ABC \), \( \angle ACB=180^{\circ}-(43^{\circ} + 58^{\circ})=79^{\circ} \).
By the Law of Sines \( \frac{AC}{\sin\angle CBA}=\frac{AB}{\sin\angle ACB} \).
Substitute \( AB = 225\), \( \angle CBA = 58^{\circ}\), \( \angle ACB=79^{\circ} \):
\( AC=\frac{225\times\sin58^{\circ}}{\sin79^{\circ}} \).
Since \( \sin58^{\circ}\approx0.848\), \( \sin79^{\circ}\approx0.982 \),
\( AC=\frac{225\times0.848}{0.982}\approx194.1\).

Step2: Find the height \( x \) using the tangent function in \( \triangle ACD \)

In right - triangle \( \triangle ACD \), \( \tan\angle CAD=\frac{x}{AC} \).
Given \( \angle CAD = 29^{\circ}\), and \( AC\approx194.1 \).
Since \( \tan29^{\circ}\approx0.554 \), then \( x = AC\times\tan29^{\circ}\).
\( x\approx194.1\times0.554\approx107.6\).

Answer:

The height \( x \) of the cliff is approximately \( 107.6\) metres.