Question

a vertical Tower stands on a horizontal plane and is surmounted by a vertical flagstaff of height 6 cm from a point the angle of elevation of the bottom and the top of the flagstaff are 45 degree and 60 degree find the height of the tower​

Answer 1

Answer:

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Answer 2

Given:

Height of flagstaff=6 cm

Angle of elevation of top and bottom of flag =60 degree and 45 degree

To find:

The height of tower

Solution:

AB=6 cm

Let height of tower=h

AC=AB+BC=h+6

In triangle BCD

$\frac{h}{CD}=tan45^{\circ}$

By using the formula

$tan\theta=\frac{perpendicular\;side}{base}$

$\frac{h}{CD}=1$

By using $tan45^{\circ}=1$

$h=CD$

In triangle ACD

$\frac{AC}{CD}=tan60^{\circ}$

$\frac{h+6}{h}=\sqrt 3$

By using $tan60^{\circ}=\sqrt 3$

$h+6=h\sqrt 3$

$6=h\sqrt 3-h=h(\sqrt 3-1)$

$h=\frac{6}{\sqrt 3-1}=\frac{6(\sqrt 3+1)}{(\sqrt 3-1)(\sqrt 3+1)}=\frac{6(\sqrt 3+1)}{3-1}=\frac{6(\sqrt 3+1)}{2}=3(\sqrt 3+1) cm$

By using the formula

$(a+b)(a-b)=a^2-b^2$

Hence,the height of tower=$3(\sqrt 3+1) cm$