Question

The angle of elevation of the top of tower from a point on the base is 60 degree and from a position which is 10 metre above the point the angle of elevation of the tower is 45 degree. Find the height of the tower and the distance of the point from the foot of the tower.
(√ 3 equal to 1.732)​

Answer 1

Answer:

Step-by-step explanation:

In ΔYRQ, we have

tan45  

o

=  

YR

QR

​  

 

1=  

YR

x

​  

 

YR=x or XP=x [because YR=XP] ---- (1)

Now In ΔXPQ we have

tan60  

o

=  

PX

PQ

​  

 

3

​  

=  

x

x+40

​  

 (from equation 1)

x(  

3

​  

−1)=40

x=  

3

​  

−1

40

​  

 

x=  

1.73−1

40

​  

=54.79 m

So, height of the tower, PQ=x+40=54.79+40=94.79 m.

Distance PX=54.79 m.

Answer 2

Answer:

ते बंगाल प्रदेशे वसन्ति। translate into English