Question

mcr3u trigonometry assignment t /29

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 angle $\angle ACB$ in $\triangle ABC$

In $\triangle ABC$, using the angle - sum property of a triangle ($\angle A+\angle B+\angle C = 180^{\circ}$).
Given $\angle CAB = 43^{\circ}$ and $\angle CBA=58^{\circ}$, then $\angle ACB=180^{\circ}-(43^{\circ}+58^{\circ}) = 79^{\circ}$.

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

The Law of Sines states that $\frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}$.
In $\triangle ABC$, let $AB = c = 225$ m, $\angle CAB = A = 43^{\circ}$, $\angle CBA=B = 58^{\circ}$, $\angle ACB=C = 79^{\circ}$. We want to find $AC$.
By the Law of Sines, $\frac{AC}{\sin\angle CBA}=\frac{AB}{\sin\angle ACB}$.
Substitute the values: $AC=\frac{AB\times\sin\angle CBA}{\sin\angle ACB}=\frac{225\times\sin58^{\circ}}{\sin79^{\circ}}$.
Since $\sin58^{\circ}\approx0.848$ and $\sin79^{\circ}\approx0.982$, then $AC=\frac{225\times0.848}{0.982}=\frac{190.8}{0.982}\approx194.3$ m.

Step3: Use the tangent function in right - triangle $\triangle ACD$

In right - triangle $\triangle ACD$, $\tan\theta=\frac{\text{opposite}}{\text{adjacent}}$.
Given $\theta = 29^{\circ}$ (angle of elevation from $A$ to $D$) and the adjacent side $AC\approx194.3$ m, and we want to find the opposite side $CD=x$.
Since $\tan29^{\circ}=\frac{x}{AC}$, then $x = AC\times\tan29^{\circ}$.
Since $\tan29^{\circ}\approx0.554$, then $x=194.3\times0.554 = 107.7$ m.

Answer:

$107.7$ m