Question

a path separates two walls. A ladder leaning against one wall rests at a point on the path. It reaches a height of 90 m on the wall and makes an angle of 60 degrees with the ground. If while resting at the same point on the path, it were made lean against the other wall, it would have made an angle of 45 degrees with the ground. Find the height it would have reached on the second wall.

Answer 1

walls= AB and CD
lenght of ladder = x
height = h

in ΔAOB 
sin 60 = AB/ AO
√3/2 = 90/x
x= 180/√3

In ΔCOD
sin 45= CD/CO
1/√2 = h/x
1/√2 = h/ 180/√3
h= 180/√6 x √6/√6
h= 30√6
h= 73.47

Answer 2

Since we do not know the length of the ladder, we need to find this. The ladder is leaning against one wall, making an angle of 60 degrees to the ground and the height at which the ladder touches the wall is 90 m.

Therefore, we can find the length of the ladder by using Sin 60 = opposite / hypotenuse = 90 / x, where x = length of the ladder.

We know that Sin 60 = (sqrt 3) / 2 = 90 / x.

Therefore, x = (90 * 2) / sqrt 3 = (30 * 2 * 3) / sqrt 3 = 60 * sqrt 3

Now that we know the length of the ladder, we can find the point at which the ladder leans on the second wall. Since it makes an angle of 45 degrees, we need to use the same formula to find the height, y.

Thus, Sin 45 = y / (60*sqrt 3)

But we know that sin 45 = 1/ (sqrt 2)

Thus, y = (60 * sqrt 3) / sqrt 2 = (30*2*sqrt3)/sqrt2 = 30*sqrt2*sqrt3 = 30*sqrt6 = 30*2.449 = 73.47m.