Question

From the top of the building of height 60m. An observer find a car at angle of depression of 30°.find how for the car is from the building.(√3=1.73)​

Answer 1

[HeY Mate]

$\huge\bold\red{Answer}$

Given

• Height of building (AM) = 60m
• Angle of depression (Θ) = 30°

To Find

• Distance between car and building (MN)

Solution

$\tt \implies \: tan 30 = \frac{perpendicular}{base} \\ \tt \implies \: \frac{1}{ \sqrt{3} } = \frac{60}{ base } \\ \tt \implies base \: = 60 \times \sqrt{3} \\ \tt \implies \: base = 60 \times 1.73 \\ \tt \implies \: base = 103.8m$

Now,

Distance between car and building = base = 103.8m

.

.

I Hope It Helps You✌️

Answer 2

${\huge{\bold{\pink{\bigstar{\boxed{\bf{Question}}}}}}}$

From the top of the building of height 60m. An observer find a car at angle of depression of 30°.find how for the car is from the building.(√3=1.73)

${\huge{\bold{\pink{\bigstar{\boxed{\bf{AnswEr}}}}}}}$

${\bold{\blue{\boxed{\bf{GivEn}}}}}$

• Height of building (AB) = 60m
• Angle of depression = 30°

${\bold{\blue{\boxed{\bf{To\: find:-}}}}}$

• Distance between car and building (BC)

${\bold{\blue{\boxed{\bf{solution}}}}}$

$\: tan 30 = \dfrac{perpendicular}{base}$

$\: \dfrac{1}{ \sqrt{3} } = \dfrac{60}{ base }$

$base \: = 60 \times \sqrt{3}$

$\: base = 60 \times 1.73$

$\: base = 103.8m$

Base = Distance btw care and building(BC) = 103.8m

_________________________❤

THANK YOU ❤