Question

the shadow of a building increases by 10 when the angle of elevation of the sun rays decreases from 60 to 45 degree what is the height of the building

please ans fast​

Answer 1

Answer:

13 m( u didnt give the unit, since it is building I took it as m)

Step-by-step explanation:

hope it helps u!!!!!

Answer 2

Answer:

15+5√3 m

Step-by-step explanation:

Let AB be the tower and let AC and AD be its shadows when the angles of elevation of the sum are 60° and 45° respectively

Then,

∠ACB=60°

∠ADB=45°

CD=10 m

Let AB=h metres and BC=x metres

In right △CAB, we have,

$\tan(60°) = \sqrt{3}$

$\tan(60°) = \frac{AB}{BC} \\ \sqrt{3 } = \frac{h}{x} \\ x = \frac{h}{ \sqrt{3} }$

x=h/√3 ......(1)

In right △ DAB, we have,

$\tan(45 ) = 1 \\ 1 = \frac{AB}{BD} \\ 1 = \frac{h}{x + 10} \\ x = h - 10$

x=h−10 .....(2)

From 1 and 2, we get,

$h - 10 = \frac{h}{ \sqrt{3} } \\ \sqrt{3} h - 10 \sqrt{3} = h \\ \sqrt{3} h - \sqrt{3} = 10 \sqrt{3} \\ h( \sqrt{3} - 1) = 10 \sqrt{3 } \\ h = \frac{10 \sqrt{3} }{ \sqrt{3 - 1} } \\ rationalising \: the \: denminator \\ h = \frac{10 \sqrt{3} }{ \sqrt{3 - 1} } \times \frac{ \sqrt{3 + 1} }{ \sqrt{3 + 1} } \\ h = \frac{10 \times 3 + 10 \sqrt{3} }{( \sqrt{3})^{2} - (1 {)}^{2} } \\ h = \frac{30 + 10 \sqrt{3} }{3 - 1} \\ h = \frac{2(15 + 5 \sqrt{3}) }{2(1)} \\ h = 15 + 5 \sqrt{3} \: m$