The angle of elevation of the top of a mobile | vedclass

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(B) Let the height of the tower $AB = h$. Let $A$ be the foot of the tower.In $\Delta PAB$,$\cot \alpha = \frac{PA}{h} \implies PA = h \cot \alpha$.In

Vedclass · Accepted Answer

$l \sin \beta$

Vedclass · Answer

(B) Let the height of the tower $AB = h$. Let $A$ be the foot of the tower.In $\Delta PAB$,$\cot \alpha = \frac{PA}{h} \implies PA = h \cot \alpha$.In $\Delta QAB$,$\cot \beta = \frac{QA}{h} \implies QA = h \cot \beta$.In $\Delta RAB$,$\cot \gamma = \frac{RA}{h} \implies RA = h \cot \gamma$.Given $\alpha : \beta : \gamma = 1 : 2 : 3$,let $\alpha = 	heta, \beta = 2	heta, \gamma = 3	heta$.$PQ = PA - QA = h(\cot 	heta - \cot 2	heta) = l$.$l = h \left( \frac{\cos 	heta}{\sin 	heta} - \frac{\cos 2	heta}{\sin 2	heta} ight) = h \left( \frac{\sin 2	heta \cos 	heta - \cos 2	heta \sin 	heta}{\sin 	heta \sin 2	heta} ight) = h \frac{\sin(2	heta - 	heta)}{\sin 	heta \sin 2	heta} = h \frac{\sin 	heta}{\sin 	heta \sin 2	heta} = \frac{h}{\sin 2	heta}$.Therefore,$h = l \sin 2	heta = l \sin \beta$.

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