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 $\angle ACB$ in $\triangle ABC$

In $\triangle ABC$, we know that the sum of angles in a triangle is $180^{\circ}$. So, $\angle ACB=180^{\circ}-\angle CAB - \angle CBA$.
Substituting the given values: $\angle ACB = 180^{\circ}-43^{\circ}-58^{\circ}=79^{\circ}$.

Step2: Use the Law of Sines in $\triangle ABC$

The Law of Sines states that $\frac{AB}{\sin\angle ACB}=\frac{AC}{\sin\angle CBA}$.
We know $AB = 225$ m, $\angle ACB = 79^{\circ}$, and $\angle CBA=58^{\circ}$.
So, $AC=\frac{AB\times\sin\angle CBA}{\sin\angle ACB}=\frac{225\times\sin58^{\circ}}{\sin79^{\circ}}$.
Using a calculator, $\sin58^{\circ}\approx0.848$ and $\sin79^{\circ}\approx0.982$.
Then $AC=\frac{225\times0.848}{0.982}=\frac{190.8}{0.982}\approx194.3$ m.

Step3: Use trigonometry in right - triangle $ACD$

In right - triangle $ACD$, $\tan\angle CAD=\frac{CD}{AC}$.
We know $\angle CAD = 29^{\circ}$ and we found $AC\approx194.3$ m.
Since $\tan\theta=\frac{\text{opposite}}{\text{adjacent}}$, and here $\tan29^{\circ}=\frac{x}{AC}$ (where $x = CD$).
So, $x=AC\times\tan29^{\circ}$.
Using a calculator, $\tan29^{\circ}\approx0.554$.
Then $x = 194.3\times0.554\approx107.8$ m.

Answer:

The height of the cliff $x$ is approximately $107.8$ m.