Question

in a square ABCD a point P is inside the square such that ABP is in equilibrium triangle the segment AP out the diagonal BD in in suppose a = the area of ABCD is​

Answer 1

Answer:

△APB is an equilateral triangle.

So, AP=PB=AB ----- (Properties of equilateral triangle)

∠PAB=∠ABP=∠BPA=60

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--- (All angles of equilateral triangle are equal and 60

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)

∠OAB=60

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-- (1)

We know that all angles all square are 90

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So, ∠CDA=90

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DB is a diagonal of square □ABCD with bisect ∠CDA equally.

Then, ∠CDB=45

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.

We know that AB=CD and AB∥CD then,

∠CDB=∠ABD=45

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-- (Alternate angles)

∠DBA=45

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i.e ∠ABO=45

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--(2)

In △AOB,

∠AOB+∠OAB+∠ABO=180

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From (1) and (2),

∠AOB+60

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+45

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=180

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∴∠AOB=75

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