Question

The diagram shows two planks are slanted on a vertical wall.
Express cot x in terms of p.

Answer 1

Solution :-

in Right ∆OAC ,

→ cot A = Base / Perpendicular

→ cot A = AC / OC

→ cot A = (1 + p) / 2 -------------- Eqn.(1)

Similarly,

in Right ∆OBC ,

→ cot B = Base / Perpendicular

→ cot B = BC / OC

→ cot B = p / 2 -------------- Eqn.(2)

Now,

→ ∠OBC = ∠OAB + ∠AOB { Exterior angle is equal to sum of interior opposite angles. }

→ ∠B = ∠A + ∠x

→ ∠x = ∠B - ∠A

Multiplying by cot both sides ,

→ cot x = cot (B - A)

using formula :-

• cot (X - Y) = [ cot X•cot Y – 1 ] / [ cot X + cot Y]

Putting value of Eqn.(1) and Eqn.(2) in RHS,

→ cot x = [(p/2)•(1+p/2) - 1] / [(p/2) + (1 + p)/2]

→ cot x = [(p + p²)/4 - 1 ] / [ (p + 1 + p)/2 ]

→ cot x = [ (p + p² - 4) / 4] / [(2p + 1) / 2]

→ cot x = [(p² + p - 4) / 4] * [(2/(2p + 1)]

→ cot x = [(p² + p - 4] / [2(2p + 1)]

→ cot x = (p² + p - 4) / (4p + 2) (Ans.)

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