Question

Example 2 : If angle B and angle Q are
acute angles such that sin B = sin Q,
then prove that angle B = angle Q.

​

Answer 1

Step-by-step explanation:

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

AB

2

=AC

2

+BC

2

(k×PR)

2

=(k×PQ)

2

+BC

2

⇒k

2

×PR

2

=k

2

×PQ

2

−BC

2

⇒BC

2

=k

2

×PR

2

−k

2

PQ

2

=k

2

[PR

2

−PQ

2

]

∴BC=

k

2

[PR

2

−PQ

2

]

From ΔPRQ

By Pythagoras theorem

PQ

2

=PR

2

+QR

2

⇒QR

2

=PQ

2

−PR

2

∴QR=

PQ

2

−PR

2

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

Hence, this is the answer