Question

From the top of a bridge that is 50 m high, two boats can be seen anchored in a marina. One boat is anchored in the direction S20°W, and its angle of depression is 40°. The other boat is anchored in the direction S60°E, and its angle of depression is 30°. Determine the distance between the two boat

Answer 1

Answer:

Explanation:

Given

height of bridge is $=50\ m$

First Boat is at an angle of $20^{\circ}$w.r.t to x axis

Second boat is at an angle of $60^{\circ}$ w.r.t to x axis

from Diagram

In triangle ABO

$\tan (40)=\frac{h}{r_1}$

$r_1=\frac{h}{\tan 40}=59.58\ m$

In triangle ACO

$r_2=\frac{h}{\tan 30}=86.602\ m$

where $r_1$ and $r_2$ are the distance of boat from origin O

Position vector of boat 1 w.r.t origin is

$\vec{x_1}=59.58\left ( -\cos (20)\hat{i}-\sin (20)\hat{j}\right )$

Position vector of boat 2 w.r.t origin is

$\vec{x_2}=86.602\left ( \cos (60)\hat{i}-\sin (60)\hat{j}\right )$

Position of $\vec{x_1} w.r.t to \vec{x_2}$

$\vec{x_{12}}=59.58\left ( -\cos (20)\hat{i}-\sin (20)\hat{j}\right )-86.602\left ( \cos (60)\hat{i}-\sin (60)\hat{j}\right )$

$\vec{x_{12}}=-99.28\hat{i}+45.209\hat{j}$

Distance between them is $|\vec{x_{12}}|=\sqrt{(-99.28)^2+(45.209)^2}$

$|\vec{x_{12}}|=109.089\ m$