Question

A straight highway leads to the foot of the tower of height 50m .From the top of the tower,the angels of depression of 2cars standing on the highway are 30° and 60°.What is the distance between the 2cars and how far is each car from the tower.​

Answer 1

${\bold{\underline{\underline{Answer:}}}}$

${\bold{\therefore Distance\:between\:two\:cars=\frac{100}{\sqrt{3}}\:m}}$

${\bold{\underline{\underline{Step-by-step\:explanation:}}}}$

• In the given question information given about a straight highway leads to the foot of the tower of height 50m .From the top of the tower,the angels of depression of 2cars standing on the highway are 30° and 60°.What is the distance between the 2cars.

• We have to find the distance between two cars.

$\underline \bold{Given : } \\ \implies Height \: of \: tower(p) = 50 \: m \\ \\ \implies 1st \: angle \: of \: depression = 60 \degree \\ \\ \implies 2nd \: angle \: of \: depression = 30 \degree \\ \\ \underline \bold{To \: Find : } \\ \implies Distance \: between \: car \: a \:and \: b = ?$

• According to given question :

$\bold{In \: \triangle \: ABC : } \\ \implies tan\: 60 \degree = \frac{p}{h} \\ \\ \implies \sqrt{3} = \frac{AB}{BC} \\ \\ \implies \sqrt{3} BC = AB \\ \\ \implies \sqrt{3} BC = \times 50 \\ \\ \bold{\implies BC= \frac{50}{ \sqrt{3} } \: m} \\ \\ \bold{In \: \triangle \: ABD : } \\ \implies tan \: 30 \degree = \frac{p}{b} \\ \\ \implies \frac{1}{\sqrt{3}} = \frac{AB}{BD} \\ \\ \implies BD= \sqrt{3} \times 50 \\ \\ \bold{\implies BD = 50\sqrt{3} \: m} \\ \\ \bold{For \: Distance \: between \: two \: cars}\\ \implies CD= BD- BC\\ \\ \implies CD = 50\sqrt{3} - \frac{50}{ \sqrt{3} } \\ \\ \implies CD = \frac{50\times 3 - 50}{ \sqrt{3} } \\ \\ \implies CD = \frac{150-50 }{ \sqrt{3} } \\ \\ \bold{\implies CD = \frac{100}{\sqrt{3}} \:m}$

Answer 2

ANSWER:-

Given:

A straight highway leads to the foot of the tower of height 50m. From the top tower, the angles of depression of 2 cars standing on the highway are 30° & 60° respectively.

To find:

What is distance between two cars & how far is each from the tower.

Solution:

We have,

•Let AB be the height of the tower= 50m

•angle of depression is 30° & 60°

•Let C & D be the position of the two car

•Let angle ADB=60° & angle ACB= 30°

Therefore,

In right ∆ABD,

[tan60° = √3]

We know, tan theta

$= > \frac{Perpendicular}{Base}$

So,

$tan60 \degree = \frac{AB}{BD} \\ \\ = > \sqrt{3} = \frac{50}{BD} \\ \\ = > \sqrt{3} BD = 50 \\ \\ = > BD = \frac{50}{ \sqrt{3} } \\ \\ = > \frac{50 \times \sqrt{3} }{ \sqrt{3} \times \sqrt{3} } \\ \\ = > \frac{50 \sqrt{3} }{3} \\ \\ = > \frac{50 \times 1.732}{3} \\ \\ = > \frac{86.6}{3} \\ \\ = >BD= 28.86m$

&

[tan30° = 1/√3]

$= > tan30 \degree = \frac{AB}{BC} \\ \\ = > \frac{1}{ \sqrt{3} } = \frac{50}{BC} \\ [cross \: multiplication] \\ = > BC = 50 \sqrt{3} \\ \\ = > BC = 50 \times 1.732 \: \: \: \: \: \: \: \: \: ( \sqrt{3} = 1.732) \\ \\ = > BC = 86.6m$

Thus,

The distance of the first car from the tower is (BD)=28.86m

&

Distance of the second car from the tower is (BC)=86.6m.

Hope it helps ☺️