Question

The angle of elevation of the top of a tower from a point A due south of the tower is 60° and from B due east of the tower is 45°. If AB = 50 m, then the height of the tower is:- (Take √3 as 1.73)
(a) 49.45 (b) 43.25 (c) 38.93 (d) 42.13

Answer 1

$\large\underline{\sf{Solution-}}$

Given that,

The angle of elevation of the top of a tower from a point A due south of the tower is 60° and from B due east of the tower is 45°.

Further given that, AB = 50 m

Let assume that CD represents the tower whose height is h meter.

Now, In right triangle BCD

$\rm \: tan45\degree = \dfrac{CD}{BC}$

$\rm \: 1 = \dfrac{h}{BC}$

$\rm\implies \:BC \: = \: h \:$

Now, In right triangle ACD

$\rm \: tan60\degree = \dfrac{CD}{AC}$

$\rm \: \sqrt{3} = \dfrac{h}{AC}$

$\rm\implies \:AC \: = \: \dfrac{h}{ \sqrt{3} }$

Now, In right triangle ABC, right angled at C

Using Pythagoras Theorem, we have

$\rm \: {AB}^{2} \: = \: {BC}^{2} + {AC}^{2}$

On substituting the values, we get

$\rm \: \dfrac{ {h}^{2} }{3} + {h}^{2} = {50}^{2}$

$\rm \: \dfrac{ {h}^{2} + 3 {h}^{2} }{3} = {50}^{2}$

$\rm \: \dfrac{4 {h}^{2} }{3} = {50}^{2}$

$\rm \: \dfrac{2{h} }{ \sqrt{3} } = {50}$

$\rm \: h = 25 \sqrt{3}$

$\rm \: h = 25 \times 1.73$

$\rm\implies \:h \: = \: 43 .25 \: m$

So, option (b) is correct.

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