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Question

Mathematics Question on measurement of angles

The angle of elevation of a cloud CC from a point P,200mP , 200 m above a still lake is 30.30^{\circ} . If the angle of depression of the image of CC in the lake from the point PP is 6060^{\circ}, then PCP C (in mm ) is equal to :

A

400

B

4003400 \sqrt{3}

C

100

D

2003200 \sqrt{3}

Answer

400

Explanation

Solution

Let PA=xP A=x For APC\triangle APC AC=PA3=x3AC =\frac{ PA }{\sqrt{3}}=\frac{ x }{\sqrt{3}} AC1=AB+BC1AC ^{1}= AB + BC ^{1} AC1=AB+BCAC ^{1}= AB + BC AC1=400+x3AC ^{1}=400+\frac{ x }{\sqrt{3}} From ΔC1PA:AC1=3PA\Delta C ^{1} PA : AC ^{1}=\sqrt{3} PA (400+x3)=3xx=(200)(3)\Rightarrow\left(400+\frac{ x }{\sqrt{3}}\right)=\sqrt{3} x \Rightarrow x =(200)(\sqrt{3}) from ΔAPC:PC=2x3PC=400\Delta APC : PC =\frac{2 x }{\sqrt{3}} \Rightarrow PC =400