The angle of elevation of the top of a tower from a point A due north of it is and from a …

Question

The angle of elevation of the top of a tower from a point $A$ due north of it is $\alpha$ and from a point $B$ at a distance of 9 units due west of $A$ is $\cos^{-1}\dfrac3{\sqrt{13}}$. If the distance of the point $B$ from the tower is 15 units, then $\cot\alpha$ is equal to:

Accepted Answer

Let $N$ be the foot of the tower, height $h$. Since angle at $B$ has $\cos^{-1}\dfrac3{\sqrt{13}}$, and $NB=15$: $\cos\theta=\dfrac{NB\text{-horizontal-component}}{NB}$... using $NA^2 = NB^2-AB^2 = 15^2-9^2=144 \implies NA=12$.

$$\tan\alpha = \frac h{NA} = \frac h{12}$$

Given the elevation-angle relation at $B$: $\dfrac h{15}=\tan\theta$ where $\cos\theta=\dfrac3{\sqrt{13}} \implies \sin\theta=\dfrac2{\sqrt{13}}\implies \tan\theta=\dfrac23$, so $h=15\times\dfrac23=10$.

$$\cot\alpha = \frac{NA}h = \frac{12}{10}=\frac65$$

Solutions · 1