Question

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❌ɴᴏ ꜱᴩᴀᴍᴍɪɴɢ ❌​

Answer 1

$\huge \mathfrak\purple{Solution}$

Let AB be the 50 m high building and CD be the tower .

In triangle CEA ,

$\frac{CE}{AE} = \tan(45°)$

$= > \frac{CE}{AE} = 1$

$= > CE = AE$ ........(i)

Also in triangle CDB ,

$\frac{CD}{BD} = \tan(60°)$

$= > \frac{CE + ED}{AE} = \sqrt{3}$(as ....AE = BD )

$= > \frac{CE + 50}{AE} = \sqrt{3}$........(ii)

Using (i) and (ii) ...

$= \frac{AE + 50}{AE} = \sqrt{3}$

$= > AE + 50 = \sqrt{3} AE$

$= > ( \sqrt{3} - 1)AE= 50$

$= > AE = \frac{50}{( \sqrt{3} - 1) }$

$= > AE = \frac{50 \times ( \sqrt{3} + 1)}{ {( \sqrt{3}) }^{2} - 1}$

$= > \frac{50 \times ( \sqrt{3} + 1) }{3 - 1}$

$= > \frac{50}{2} ( \sqrt{3} + 1)$

$= > 25(1.73 + 1)$

$= > 25 \times 2.73$

$= > 68.25$

$AE= > 68.25m$

By eq (i) .....

CE = 68.25 m

Height of tower = CE + ED

= 68.25 + 50

= 118.25 m

Hence , the height of the tower is 118.25 m and the horizontal distance between the tower and the building is 68.25 m.

Answer 2

Answer :

• Height of the Pole is 50/√3 m

• Horizontal Distance between Tower and Pole is 50√3 m

Explanation :

Kindly Check out the Attachment

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