Question

If angleB and angleQ are acute angles such that sin B=sin Q, then prove that angle B= angle Q.(with explanation)

Answer 1

Given that ∠B and ∠Q are acute angle and

sinB=sinQ__ (A)

From ΔACB and ΔPRQ

sinB= AB/AC __(1)

sinQ= PQ/PR ___(2)

From equation (A)

sinB=sinQ

AB/AC = PQ/PR

let AB/AC = PQ/PR =k

∴ PR/AC = PQ/AB =k __(3)

Now,

AC=k×PR

AB=k×PQ

From ΔACB

By Pythagoras theorem

AB2 =AC2+BC2

(k×PR)2 =(k×PQ)2 +BC2

⇒k2 ×PR2 =k2 ×PQ2 −BC2

⇒BC2 =k2 ×PR2 −k2 PQ2

=k2 [PR2 −PQ2 ]

∴BC= √k2[PR2 −PQ2 ]

From ΔPRQ

By Pythagoras theorem

PQ2 =PR2 +QR2

⇒QR2 =PQ2 −PR2

∴QR= √PQ2 −PR2

Consider that

QR/BC =k __(4)

From equation (3) and (4) to,

PR/AC = PQ/AB = QR/BC

Hence, ΔACB∼ΔPRQ (sss similarity)

∠B=∠Q

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