Question

A straight highway leads to the foot of the tower of height 50m.from the top of the tower,the angle 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

Answer:

Step-by-step explanation:

Let AB be the tower of height 50 m.

Let C and D be the position of the 2 cars.

Let ∠ADB = 60° 

and ∠ACB = 30°

In ∆ABD.

tan 60° = AB / BD

under root 3 = 50 / BD

BD = 50 / under root 3

50 underroot 3 /3

50 ×1.732 / 3

86.6 / 3 = 28.86m.

In triangle ABC ,

Tan 30° = AB / BC

1 / underroot 3 = 50/BC

BC = 50 underroot 3 = 86.6 m.

the distance of the first car from the tower =》28.86m

distance of second car = 86.6m

Answer 2

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

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

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

• 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.

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

• According to given question :

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