Question

The tallest tower in a city is 100 metre high and a multistoried hotel at the City Centre is 20 m high. The angle of elevation of the top of a tower at the top of the hotel is 30 degree. A building 'h' metre high is situated on road connecting the tower with the City Centre at the distance of 1 km from the Tower. find the value of h if the top of the tower , building and the hotels are in the same straight line.

Answer 1

I HAVE TAKEN THE DISTANCE AS 10 M AND NOT 1 KM.
PLEASE CLARIFY WITH VALUES.
We would get a diagram
Imagine AB is the tallest tower.
CD is the city center.
And if you draw a straight line to AB from C it is X.
So angle XAC = 30 degree.
Now join A to C.
We have MN as the building touching AC at N.
So BM is 1 m.
Let MN intersect XC at O
Now BX = CD = 20m hence AX = 100 - 20 = 80 m
Now in triangle XCA
Tan 30 = AX / XC
1/3^2 = 80/10 + OC  {XC =BD}
OC = 80 (3)^2 - 10
Now in Triangle OCN
Tan 30 = ON / OC
1/3^2 = ON / 80(3)^2 - 10
ON = 74.21 m ( by solving the above equation )
Height of tower MN is ON + OM = 74.21 + 20 = 94.21 m
Hence 94.21 m
HOPE IT HELPS!