2. sheryl is surveying a cliff to determine the elevation from the base of a canyon to the top of the cliff…

2. 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

2. 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

Answer

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.