A man observes that when he moves up a distance c metres on a slope, the angle of depression of a point on the horizontal plane from the base of the slope is 30°, and when he moves up further a distance c metres, the angle of depression of that point is 45°. The angle of inclination of the slope with the horizontal is?


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Question Number 15393 by Tinkutara last updated on 10/Jun/17

$A man observes that when he moves up$ $a distance c metres on a slope, the$ $angle of depression of a point on the$ $horizontal plane from the base of the$ $slope is 30 °, and when he moves up$ $further a distance c metres, the angle of$ $depression of that point is 45 ° . The$ $angle of inclination of the slope with the$ $horizontal is ?$
	Answered by mrW1 last updated on 10/Jun/17

	$let$ $a = distance of the point to foot of slope$ $α = angle of slope$ $γ = \frac{a}{c}$ $(a + c \cos α) \tan 30 ° = c \sin α$ $(γ + \cos α) \frac{1}{\sqrt{3}} = \sin α . . . (i)$ $(a + 2 c \cos α) \tan 45 ° = 2 c \sin α$ $(γ + 2 \cos α) = 2 \sin α . . . (ii)$ $(ii) / (i) :$ $\frac{γ + 2 \cos α}{γ + \cos α} \times \sqrt{3} = 2$ $γ \sqrt{3} + 2 \sqrt{3} \cos α = 2 γ + 2 \cos α$ $2 (\sqrt{3} - 1) \cos α = (2 - \sqrt{3}) γ$ $2 (\sqrt{3} - 1) \cos α = 2 (2 - \sqrt{3}) (\sin α - \cos α)$ $\cos α = (2 - \sqrt{3}) \sin α$ $\tan α = \frac{1}{2 - \sqrt{3}} = 2 + \sqrt{3}$ $α = \tan^{- 1} (2 + \sqrt{3}) = 75 °$
	Commented by mrW1 last updated on 10/Jun/17

	Commented by Tinkutara last updated on 10/Jun/17

	$Thanks Sir!$
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