Question

find angle PCQ if arch side of the square is 1 unit and perimeter of ∆AQP=2units ​

Answer 1

Answer:

Your answer is 45°

Step-by-step explanation:

Since the perimeter of the right triangle △APQ is 2 units, we have

PQ = 2−p−q

Applying Pythagoras theorem on △APQ,

p^2+q^2 = (2−p−q)^2

⟹q = 2(1−p)/2−p -------------------------------------------------->  (1)

Applying Pythagoras theorem on △PBC,

PC = root (PB^2 + BC^2) = root (1−p)^2 + 1^2

Applying Pythagoras theorem on △QBC, we similarly get

QC = root ( q^2- 2q + 2 )

Applying cosine rule on ∠PCQ, we have

∠PCQ = arc cos[ (PQ^2−PC^2−QC^2) / 2⋅PC⋅QC ]

           =arc cos[ { [(2−p−q)^2−(p^2−2p+2)−(q^2−2q+2)] / 2⋅p^2−2p+2} ]

                                DIVIDED BY

              2 ⋅ root (p^2−2p+2) . root (q^2−2q+2)

Substituting q from (1) and simplifying, we get

    ∠PCQ = arc cos (1 / root 2)

               = $\pi$/4 or 45°