a man notices two objects in a straight line due west after walking a distance c due north he - Maths - Some Applications of Trigonometry

Question

a man notices two objects in a straight line due west after
walking a distance c due north he observes that the objects subtend
an angle A at his eye; and after walking a further distance 2c due
north they subtend an angle B if height of the man is being
neglected then prove that the distance between the objects is
8c / 3cotB - cotA

Koka Sri Lakshmi Divya Sai · Accepted Answer

$Let a man observe two objects P and Q due west of him at O R and S are two positions of the man due north of O, such that OR=c and RS=2c ∠ P R Q = A and ∠ PSQ=B Let PQ=x and OP=y In right triangle POR and QOR,we have tan ∠ ORP = \frac{y}{O R} \tan ∠ O R Q = \frac{x + y}{O R} But A= ∠ O R Q - ∠ O R P ∴ \tan A = \tan (∠ O R Q - ∠ O R P) = \frac{\tan ∠ O R Q - \tan ∠ O R P}{1 + \tan ∠ O R Q \tan ∠ O R P} = \frac{(\frac{x + y}{O R}) - (\frac{y}{O R})}{1 + (\frac{x + y}{O R}) (\frac{y}{O R})} \Rightarrow \tan A = \frac{x . O R}{O R^{2} + x y + y^{2}} \Rightarrow x y + y^{2} + O R^{2} = x . O R \cot A - - - - (1) Similarly tanB=tan (∠ O S Q - ∠ O S P) \Rightarrow \tan B = \frac{\frac{(x + y)}{O S} - \frac{y}{O S}}{1 + (\frac{x + y}{O S}) (\frac{y}{O S})} \Rightarrow x y + y^{2} + O S^{2} = x . O S \cot B - - - (2) Subtracting (1) from (2) we get {OS}^{2} - O R^{2} = x (O S \cot B - O P \cot A) \Rightarrow (3 c)^{2} - c^{2} = x (3 c \cot B - c \cot A) \Rightarrow x = \frac{8 c}{3 \cot B - \cot A}$

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