waecmaths question:
A stationary boat is observed from a height of 100 m . If the horizontal distance between the observer and the boat is 80 m. Calculate, correct to two decimal places, the angle of depression of the boat from the point of observation
Option A:
36.87o
Option B:
39.70o
Option C:
51.13o
Option D:
53.13o
waecmaths solution:
$\begin{align} & \text{Let }\angle ACD=\theta \text{ be the angle of depression} \\ & \text{From geometry} \\ & \angle CAB=\angle ACD\{Alternate\text{ }angles\} \\ & consider\text{ }\vartriangle ABC \\ & \tan \theta =\frac{BC}{AB}=\frac{100}{80}=1.25 \\ & \theta ={{\tan }^{-1}}1.25=\text{51}\text{.3}{{\text{4}}^{\circ }} \\\end{align}$
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