Question

Two men on either side of 75m high building and in line with Base of building observe the angle of elevation of the top of the building as 30 and 60 .find the distance between the two men

Answer 1

Answer:

$\red { Distance \: between \:two \:men }$

$\green {= 173.2 \:m}$

Step-by-step explanation:

From the figure ,

Height of the building = AD = 75 m ,

Let Distance from the foot of the building to the one observer = BD

Distance from the foot of the building to the second observer = DC

$\angle ABD = 60\degree$

$\angle ACD = 30\degree$

$In \: \triangle ADB ,\\tan\: \angle 60 = \frac{AD}{BD}$

$\implies \sqrt{3} = \frac{75}{BD}$

$\implies BD = \frac{75}{sqrt{3}}\\= \frac{75\sqrt{3}}{\sqrt{3}\times \sqrt{3}}\\= \frac{75\sqrt{3}}{3} \\= 25\sqrt{3} \:m\:---(1)$

$In \: \triangle ADC ,\\tan\: \angle 30 = \frac{AD}{DC}$

$\implies \frac{1}{\sqrt{3}}= \frac{75}{DC}$

$\implies DC = 75\sqrt{3}\:--(2)$

Therefore.,

$\red { Distance \: between \:two \:men }$

$= BD+DC$

$= 25\sqrt{3} + 75\sqrt{3}\\= 100\sqrt{3}\\=100\times 1.732$

$\green {= 173.2 \:m}$

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Answer 2

Step-by-step explanation:

refer to second pic everything is explained properly you can ask doubt in comments and pls like this answer