Question

Angle of elevation of top of a tower from point x from the ground is 60° . From a point 40m vertically above x, the angle of depression of base of tower is 30° . Find height of the tower.​

Answer 1

Answer:

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

Step-by-step explanation:

hajj hhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhivigifififigydtststdvjvjvjvuvufugivkbkbkvkvk hhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhivigifififigydtststdvjvjvjvuvufugivkbkbkvkvk