41. AB is a vertical pole. The end A is on theground, Cis the middle p

1.	41. AB is a vertical pole. The end A is on theground, Cis the middle point of AB and P is apoint on the level ground. The portion BCsubtends an angle α at P. If AP= n. AB, then Itan αis:
Answer» AB is vertical pole with A at ground. C is midpoint of AB P is another point in ground and Distance of AP= nAB Let AB=X So, AC=X/2 AP= nX Given that Beta is the angle subtended by BC at P Assume Alpha is the angle subtended by AC, such that AB subtends an angle of (Alpha +Beta) at P Therefore, tan (Alpha) = (X/2) / nX = 1/2n tan (Alpha +Beta) = X/nX = 1/n We know the formula: tan (x+y) = [tan x + tan y]/ [1- tan x * tan y] Therefore, 1/n= [tan Beta + 1/2n ] / [ 1- (tan beta*1/2n)] After solving the above equation, tan Beta = n/(2n^2+1)

41. AB is a vertical pole. The end A is on theground, Cis the middle point of AB and P is apoint on the level ground. The portion BCsubtends an angle α at P. If AP= n. AB, then Itan αis:

Answer»

AB is vertical pole with A at ground. C is midpoint of AB

P is another point in ground and Distance of AP= nAB

Let AB=X

So, AC=X/2

AP= nX

Given that Beta is the angle subtended by BC at P

Assume Alpha is the angle subtended by AC, such that AB subtends an angle of (Alpha +Beta) at P

Therefore, tan (Alpha) = (X/2) / nX = 1/2n

tan (Alpha +Beta) = X/nX = 1/n

We know the formula: tan (x+y) = [tan x + tan y]/ [1- tan x * tan y]

Therefore, 1/n= [tan Beta + 1/2n ] / [ 1- (tan beta*1/2n)]

After solving the above equation,

tan Beta = n/(2n^2+1)