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2015-03-06T15:53:49+05:30

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height of mountain = OA  =H

In ΔAOB,
tan (ABO) = AO/OB
⇒ tan 45 = AO/OB
⇒ 1 = AO/OB
⇒ OB = AO = H

In ΔPBR,
sin 30 = PR/PB
⇒ 1/2 = y/1000
⇒ y = 1000×(1/2) = 500m
Thus, AQ = AO - OQ = (H-500)m

cos 30 = RB/PB
⇒ √3 / 2 = x/1000
⇒ BR = 1000×(√3 / 2)
⇒ BR = 500√3m
thus OR = BO - BR = (H - 500√3)m
thus PQ = (H-500√3)m   (since OP=PQ)

In ΔAPQ,

tan\ 60= \frac{AQ}{PQ}\\ \\ \Rightarrow \sqrt{3} = \frac{H-500}{H-500 \sqrt{3} } \\ \\ \Rightarrow \sqrt{3}(H-500 \sqrt{3})=H-500 \\ \\ \Rightarrow  \sqrt{3}H-1500=H-500\\ \\\Rightarrow  \sqrt{3}H - H = -500+1500 \\ \\ \Rightarrow H( \sqrt{3}-1)=1000\\ \\ \Rightarrow H= \frac{1000}{\sqrt{3}-1}=\frac{1000}{\sqrt{3}-1} \times  \frac{\sqrt{3}+1}{\sqrt{3}+1} \\ \\ \Rightarrow H= \frac{1000(\sqrt{3}+1)}{2}=500(\sqrt{3}+1)\ m
2 5 2
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