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(A) Let the height of the tower be $h$ and the distance of the first car from the tower be $x_1 = 75\, m$. Let the distance of the second car be $x_2$

Vedclass · Accepted Answer

$150$

Vedclass · Answer

(A) Let the height of the tower be $h$ and the distance of the first car from the tower be $x_1 = 75\, m$. Let the distance of the second car be $x_2$. The angles of depression are $60^{\circ}$ and $30^{\circ}$,which correspond to the angles of elevation from the cars to the top of the tower.In the right-angled triangle formed by the first car,the tower base,and the top of the tower: $	an(60^{\circ}) = h / 75$. Thus,$h = 75 	imes \sqrt{3} = 75\sqrt{3}\, m$.In the right-angled triangle formed by the second car,the tower base,and the top of the tower: $	an(30^{\circ}) = h / x_2$. Thus,$x_2 = h / 	an(30^{\circ}) = (75\sqrt{3}) / (1 / \sqrt{3}) = 75 	imes 3 = 225\, m$.The distance between the two cars is $x_2 - x_1 = 225 - 75 = 150\, m$.

Two cars are parked in the same direction of a tower. Observing from the top of the tower,their angles of depression are found to be $60^{\circ}$ and $30^{\circ}$. If the car nearer to the tower is $75\, m$ away from the tower,find the distance between these two cars (in $m$).

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