The shadow of tower- when the angle of elevation of the sun is 45-o- is found to be 10m longer than when it was 60-o- The height of the tower is approx-8-66m21m31m24m

Question

Toppr Admin · Accepted Answer

The correct option is A 8-66mGiven d-10m-x3B8-1-45-x2218-x3B8-2-60-x2218-Formula is height of the tower -h- d-cot-x3B8-1-x2212-cot-x3B8-2-Put the values in the above formula- we get height of the tower is 8-66m

The shadow of tower, when the angle of elevation of the sun is $45^{o}$ , is found to be $10$ m longer than when it was $60^{o}$ . The height of the tower is approx.
$8.66$ m
$21$ m
$31$ m
$24$ m

The correct option is A $8.66$ m
Given
$d = 10 m$
$θ_{1} = 45^{\circ}$
$θ_{2} = 60^{\circ}$
Formula is height of the tower $(h) =$ $\frac{d}{(cot θ_{1} - cot θ_{2})}$
Put the values in the above formula, we get height of the tower is $8.66 m$

The shadow of tower, when the angle of elevation of the sun is 45o, is found to be 10m longer than when it was 60o. The height of the tower is approx.8.66m21m31m24m

The correct option is A 8.66mGiven d=10mθ1=45∘θ2=60∘Formula is height of the tower (h)= d(cotθ1−cotθ2)Put the values in the above formula, we get height of the tower is 8.66m

The shadow of tower, when the angle of elevation of the sun is $45^{o}$ , is found to be $10$ m longer than when it was $60^{o}$ . The height of the tower is approx.
$8.66$ m
$21$ m
$31$ m
$24$ m

The correct option is A $8.66$ m
Given
$d = 10 m$
$θ_{1} = 45^{\circ}$
$θ_{2} = 60^{\circ}$
Formula is height of the tower $(h) =$ $\frac{d}{(cot θ_{1} - cot θ_{2})}$
Put the values in the above formula, we get height of the tower is $8.66 m$