Answer to Question #165326 in Trigonometry for Surendra Singh

Question

A ladder makes alpha angle with the ground. Its bottom is pulled by a And due to this it slips by b with the wall and makes beta angle with the wall then a is equal to

a. b tan(alpha+beta)/2

b. b tan(alpha+beta)

c. b tan( ( alpha+beta)/2)

d. (b tan((alpha+beta)/2))/2

Answer 1

Worth noting, the data in problem don't correspond to the answers. In answers the two angles are symmetric, so $\alpha$ and $\beta$ should be the angles with the ground.

Let l be the length of the ladder. Therefore, the distance of the bottom from the wall is $l\cos\alpha$ and the height of top is $l\sin\alpha.$ If we pull the bottom by a, the distance from the wall will be $l\cos\beta= l\cos\alpha + a$ and the height of the top will be $l\sin\beta =l\sin\alpha - b.$

Therefore,

$l\cos\beta= l\cos\alpha + a, \\ l\sin\beta =l\sin\alpha - b.$

So $l = \dfrac{b}{\sin\alpha - \sin\beta}$ and $a =b\cdot \dfrac{\cos\beta-\cos\alpha}{\sin\alpha - \sin\beta} = b\cdot\dfrac{-2\sin\frac{\alpha+\beta}{2}\sin\frac{-\alpha+\beta}{2}}{2\sin\frac{\alpha-\beta}{2}\cos\frac{\alpha+\beta}{2}} =b \tan\dfrac{\alpha+\beta}{2}$ .