The angle of elevation of the top of a tower | vedclass

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(A) Let the tower be $OP$,where $O$ is the base of the tower. Point $A$ is due north of $O$,so $	riangle OAP$ is a right-angled triangle with $\angle

Vedclass · Accepted Answer

$\frac{6}{5}$

Vedclass · Answer

(A) Let the tower be $OP$,where $O$ is the base of the tower. Point $A$ is due north of $O$,so $	riangle OAP$ is a right-angled triangle with $\angle OAP = \alpha$. Given $AB = 9$ units,where $B$ is due west of $A$. Since $A$ is north of $O$,$OA \perp AB$. In $	riangle OAB$,$\angle OAB = 90^\circ$. Given $OB = 15$,by Pythagoras theorem,$OA^2 + AB^2 = OB^2 \implies OA^2 + 9^2 = 15^2 \implies OA^2 = 225 - 81 = 144 \implies OA = 12$. Let $\beta = \cos^{-1}\left(\frac{3}{\sqrt{13}}ight)$ be the angle of elevation from $B$. Then $\cos \beta = \frac{3}{\sqrt{13}}$. Since $\sin^2 \beta + \cos^2 \beta = 1$,$\sin \beta = \sqrt{1 - \frac{9}{13}} = \sqrt{\frac{4}{13}} = \frac{2}{\sqrt{13}}$. Thus,$	an \beta = \frac{\sin \beta}{\cos \beta} = \frac{2/\sqrt{13}}{3/\sqrt{13}} = \frac{2}{3}$. In $	riangle OBP$,$	an \beta = \frac{OP}{OB} = \frac{h}{15}$. So,$\frac{h}{15} = \frac{2}{3} \implies h = 10$. In $	riangle OAP$,$	an \alpha = \frac{OP}{OA} = \frac{h}{OA} = \frac{10}{12} = \frac{5}{6}$. Therefore,$\cot \alpha = \frac{1}{	an \alpha} = \frac{6}{5}$.

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 $\cos^{-1}\left(\frac{3}{\sqrt{13}}\right)$. If the distance of the point $B$ from the tower is $15$ units,then $\cot \alpha$ is equal to.

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