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 \\( \angle cab \\) to be \\( 43 ^ { \circ } \\), \\( \angle cba \\) to be \\( 58 ^ { \circ } \\), and the angle of elevation from point a to point d to be \\( 29 ^ { \circ } \\). solve for the height x of the cliff, to the nearest tenth of a metre. 5t

Explanation:

Step1: Find the length of \( AC \)

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

Step2: Find the height \( x \)

In \(\triangle ACD\), \(\sin\angle CAD=\frac{x}{AC}\). Given \(\angle CAD = 29^{\circ}\), then \(x = AC\times\sin29^{\circ}\).
Since \(AC\approx194.1\) m and \(\sin29^{\circ}\approx0.485\), \(x=194.1\times0.485\approx94.1\) m.

Answer:

\(94.1\) m