The angle of elevation of the top of a tower from a point on the ground is 30 degree

Text Solution

Solution : Let ADC be traingle right angled at C<br> DC be the height and AC bee the base<br> B is the mid point of AC<br> Length of AB=20m<br> Length of BC=d-20 m<br> Length of AC(AB+BC)=d m<br> Also,`/_B=60^0`<br> `/_A=30^0`<br> so,<br> In `DeltaADC tan30^0=(DC)/(AC)​ `<br> `1/sqrt(3)​=h/d`​ ........(1) <br> `d=sqrt(3) ​h`<br> `d=sqrt(3) xx10sqrt(3) ​ =30m`<br> In `DeltaBCD tan60^0=(DC)/(BC)​ `<br> `sqrt(3)=h/(d−20)​`.......(2)<br> on solving equation (1) & (2)<br> `sqrt(3)=h/(sqrt(3h−20)`<br> ​ `3h−20sqrt(3) ​=h`<br> ​`h=10sqrt(3) ​m`<br> height of tower.

Uh-Oh! That’s all you get for now.

We would love to personalise your learning journey. Sign Up to explore more.

Sign Up or Login

Skip for now

Uh-Oh! That’s all you get for now.

We would love to personalise your learning journey. Sign Up to explore more.

Sign Up or Login

Skip for now

Uh-Oh! That’s all you get for now.

We would love to personalise your learning journey. Sign Up to explore more.

Sign Up or Login

Skip for now

Uh-Oh! That’s all you get for now.

We would love to personalise your learning journey. Sign Up to explore more.

Sign Up or Login

Skip for now