Question

From the top of a vertical tower the angle of depression of two cars in the same straight line with the base of the tower at an instant/are found to be 45° and 60° if the cars are 100m apart and are on the same side of the tower find the height of tower.​

Answer 1

Answer:

236.6 meters

Step-by-step explanation:

Let OP be the tower and points A and B be the positions of the cars.

Now, AB=100 m, ∠OAP=60° , ∠OBP=45°

Let OP=h

In △AOP,

tan60° =  ​OP/OA

√3 ​ = h/OA

OA = h/√3 ​

Also, in △BOP,

tan45 = OP/OB

1 = h/OB

OB= h

Now,

OB−OA=100

h − h/√3 =100

h( √3 − 1) ​ =100

h = 100√3/√3 - 1​​

h=236.6 m

hence, the height of the tower is 236.6 m.

Hope it helps you :^)

Answer 2

Answer:

$\bf \fbox \red { the height of tower. is 236.59 metre}$

Step-by-step explanation:

THE HEIGHT OF THE TOWER IS AB

In ∆ABC ,

$\tan( {60}^{0} ) = \frac{AB}{x} \\ \\ x \sqrt{3} = AB...........(eq \: 1)$

________________________

In ∆ABD

$\tan( {45}^{0} ) = \frac{AB}{100 + x} \\ \\ 100 + x = AB..............(eq \: 2)$

________________________

From equation 1 & 2

$100 + x = x \sqrt{3} \\ \\ 100 = x\sqrt{3} - x \\ \\ 100 = x( \sqrt{3} - 1) \\ \\ 100 = x(1.732 - 1) \\ \\ 100 = x \times 0.732 \\ \\ \frac{100}{0.732} = x \\ \\ 136.6 \: m = x$

________________________

In ∆ABC

$\tan( {60}^{0} ) = \frac{AB}{x} \\ \\ x \sqrt{3} = AB$

_______________________

Substitute the value of x = 136.6

Therefore,

136.6 x √3 = AB

136.6 x 1.732 = AB

236.59 metre = AB

The height of the tower is 236.59 metre.