Question

On the same side of a tower, two objects are located. When observed from the top of the tower, their angles of depression are 45° and 60°. If the height of the tower is 150 m, find the distance between the objects.

Answer 1

Answer:

The  distance between the objects is 63.4 m.

Step-by-step explanation:

Given :  

Height the tower ,AB =  150 m and Two objects C & D from the top of the tower  makes an angle of depression from the top and bottom of tower are 45° and 60°.  

Let CD = 'x' m , BD = 'y' m and ∠ADB = 60°,   ∠ACB = 45°

In right angle ∆ABD,

tan θ  = P/ B

tan 60° = AB/BD

√3 =  150/y

√3 = 150/y

√3 × y = 150  

y = 150/√3 ……………(1)

In right angle ∆ABC,

tan θ  = P/ B

tan 45° = AB/BC

tan 45° = AB/(BD + DC)

tan 45° = 150/(x + y)

1 = 150/(x + y)

x + y = 150

x + 150/√3 = 150

[From eq 1]

x = 150 - 150/√3

x = 150( 1 - /1√3)

x = 150(√3 - 1)/√3

√3x = 150(√3 - 1)

1.732 x = 150(1.732 - 1)

[√3 = 1.732]

1.732 x = 150(0.732)

1.732 x = 109.8

x = 109.8/1.732  

x = 63.4 m

Hence , the  distance between the objects is 63.4 m.

HOPE THIS ANSWER WILL HELP YOU…

Answer 2

$\huge\bigstar\mathfrak\purple{\underline{\underline{SOLUTION:}}}$

In right ∆ABC,

$tan(60 \degree) = \frac{</strong><strong>AB</strong><strong>}{</strong><strong>BC</strong><strong>} \\ \\ = > \sqrt{3} = \frac{150}{</strong><strong>BC</strong><strong>} \\ \\ = > </strong><strong>BC</strong><strong>= \frac{150}{ \sqrt{3} } = 50 \sqrt{3}$

In right ∆ABD,

$tan(45 \degree) = \frac{</strong><strong>AB</strong><strong>}{</strong><strong>BD</strong><strong>} \\ \\ = > 1 = \frac{150}{</strong><strong>BD</strong><strong>} \\ \\ = > </strong><strong>BD</strong><strong> = 150m$

Therefore,

Distance between the 2 objects,

CD= BD - BC

=) 150 - 50√3

=) 63.4m

hope it helps ☺️