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Question: If from the top of tower of \(60\) metre height, the angles of depression of the top and floor of a ...

If from the top of tower of 6060 metre height, the angles of depression of the top and floor of a house are α,β\alpha ,\beta respectively and if the height of the house is 60sin⁡(β−α)x\dfrac{{60\sin \left( {\beta - \alpha } \right)}}{x}, then xx is
A. sin⁡αsin⁡β\sin \alpha \sin \beta
B. cos⁡αcos⁡β\cos \alpha \cos \beta
C. sin⁡αcos⁡β\sin \alpha \cos \beta
D. cos⁡αsin⁡β\cos \alpha \sin \beta

Explanation

Solution

Choose different triangles and apply the formula of trigonometric ratios of sine, cosine and tangent. Simplify the trigonometric ratio and find the value of the height. Use the concept of triangles to get the answer.

Complete step-by-step answer:
Consider the diagram

In triangle ABDABD,
tanβ=DBAB{{tan\beta }} = \dfrac{{{\text{DB}}}}{{{\text{AB}}}}
tanβ=60d{{tan\beta }} = \dfrac{{{\text{60}}}}{{\text{d}}}
d=60  cotβ          (∴  1tan⁡β=cot⁡β)......(1)d = 60\;{{cot\beta }}\;\;\;\;\;\left( {\therefore \;\dfrac{1}{{\tan \beta }} = \cot \beta } \right)......\left( 1 \right)
Now in triangle DECDEC,
tan⁡α=DEEC\tan \alpha = \dfrac{{DE}}{{EC}}
⇒\Rightarrow tan⁡α=DCd\tan \alpha = \dfrac{{DC}}{d}
⇒\Rightarrow DC=dtan⁡α......(2)DC = d\tan \alpha ......\left( 2 \right)
Here, DC=60−hDC = 60 - h because DCDC is calculated by subtracting DB=hDB = h.
Substitute the value of DC=60−hDC = 60 - h and dd from equation (1).
Therefore,
⇒\Rightarrow 60−h=60cot⁡βtan⁡α......(3)60 - h = 60\cot \beta \tan \alpha ......\left( 3 \right)
Now on simplifying equation (3).
⇒h=60(1−cos⁡βsin⁡β⋅sin⁡αcos⁡α)           (∴  cot⁡β=cos⁡βsin⁡β,  tan⁡α=sin⁡αcos⁡α)   \Rightarrow h = 60\left( {1 - \dfrac{{\cos \beta }}{{\sin \beta }} \cdot \dfrac{{\sin \alpha }}{{\cos \alpha }}} \right)\;\;\;\;\,\;\left( {\therefore \;\cot \beta = \dfrac{{\cos \beta }}{{\sin \beta }},\;\tan \alpha = \dfrac{{\sin \alpha }}{{\cos \alpha }}} \right) \\\ \\\
Take LCM and simplify it becomes,
⇒\Rightarrow h=60sin⁡(β−α)cos⁡αsin⁡βh = \dfrac{{60\sin \left( {\beta - \alpha } \right)}}{{\cos \alpha \sin \beta }}
⇒\Rightarrow h=60sin⁡(β−α)x           (  given  cos⁡αsin⁡β=x)h = \dfrac{{60\sin \left( {\beta - \alpha } \right)}}{x}\,\;\;\;\;\;\left( {\;given\;\cos \alpha \sin \beta = x} \right)
Therefore,
60sin⁡(β−α)x=60sin⁡(β−α)cos⁡αsin⁡β\dfrac{{60\sin \left( {\beta - \alpha } \right)}}{x} = \dfrac{{60\sin \left( {\beta - \alpha } \right)}}{{\cos \alpha \sin \beta }}
⇒\Rightarrow x=cos⁡αsin⁡βx = \cos \alpha \sin \beta

Hence, the option D is correct.

Note: Euclid and Archimedes presented the concept of Trigonometry. Trigonometry is a branch of mathematics that studies between side lengths and angles. Trigonometry helps to solve the problems of right-angled triangles.
Trigonometry ratios are the ratios between edges of the right-angle triangle. There are six trigonometric ratios sin, cos, tan, cosec, sec, cot. Sine function defined as the ratio of perpendicular to the hypotenuse. Cos function is defined as the ratio of base to the hypotenuse. Tan function is defined as the ratio of perpendicular to the base. The reciprocal of these functions defines cosec, sec and cot respectively.