AB is vertical tower. The point A is on the ground and C is the middle point of AB. The part CB subtend
an angle a at a point P on the ground. If AP = nAB, then tan a =
(1) n(n² + 1)
(2)n/(2n²-1)
(3) n²/(2n²+1)
(4)n/(2n²+1)
Answers
Answer:
Given:
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=2X
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 (α)=nX(X/2)=2n1
tan(α+β)=nXX=n1
We know the formula: tan(x+y) = [1−tanx∗tany][tanx+tany]
Therefore, n1= [1−(tanβ∗1/2n)][tanβ+1/2n]
After solving the above equation,
tanβ=(2n2+1)n
Answer:
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