A $15 m$ long ladder touches the top of vertical wall. If this ladder makes an angle of $60^{0}$ with the wall, then height of the wall is :

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Solution

REF.Image
$s i n 60^{\circ} = \frac{A B}{A C} = \frac{A B}{15}$
$A B = 15 s i n 60^{\circ} = \frac{15 \sqrt{3}}{2} m$

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Q. Choose the correct answer of the following question:

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A 15m long ladder touches the top of vertical wall. If this ladder makes an angle of 600 with the wall, then height of the wall is :

REF.Imagesin60∘=ABAC=AB15AB=15sin60∘=15√32m

A $15 m$ long ladder touches the top of vertical wall. If this ladder makes an angle of $60^{0}$ with the wall, then height of the wall is :

REF.Image
$s i n 60^{\circ} = \frac{A B}{A C} = \frac{A B}{15}$
$A B = 15 s i n 60^{\circ} = \frac{15 \sqrt{3}}{2} m$