Question

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If angleB and angleQ are acute angles such that sinB = sinQ, then prove that angleB = angleQ.

Figure : Attached above.

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Answer 1

Given:-

• angleB and angleQ are acute angles such that sinB = sinQ

To prove :-

prove that angleB = angleQ

Proof:-

Let us suppose that ABQ is a right angle such that triangle is right angled at B

Given, SinB = SinQ

Now,

$\sf{:\implies SinB=\dfrac{AQ}{BQ}.....(I)}$

$\sf{:\implies SinQ=\dfrac{AB}{BQ}.....(ii)}$

Since, SinB = SinQ

then, from eq (I) and (ii) we get,

$\sf{:\implies \dfrac{AQ}{BQ}=\dfrac{AB}{BQ}}$

$\sf{:\implies AQ=AB}$

Therefore, triangle ABC is an isosceles triangle.

.°. Angle B = Angle C

Hence, Proved ............

Answer 2

Answer:

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Step-by-step explanation:

Given that ∠B and ∠Q are acute angle and

sinB=sinQ__ (A)

From ΔACB and ΔPRQ

sinB= Ac/AB. __(1)

sinQ= PR/PQ ___(2)

From equation (A)

sinB=sinQ

Ac/AB.=PR/PQ=k

AC/PR = AB/PQ =k __(3)

Now,

AC=k×PR

AB=k×PQ

From ΔACB

By Pythagoras theorem

AB² =AC² +BC ²

(k×PR) ²=(k×PQ) ² +BC²

⇒k ² ×PR ² =k ²×PQ ² −BC ²

⇒BC²=k ² ×PR ²−k ²PQ²

=k ² [PR ² −PQ ²)

∴BC=√k² [PR ²−PQ ² ]

From ΔPRQ

By Pythagoras theorem

PQ ²=PR² +QR ²

⇒QR² =PQ ² −PR ²

∴QR= √PQ ² −PR ²

Consider that

BC/QQ =k. __(4)

From equation (3) and (4) to,

AC/ PR = AB/PQ = BC/QR

Hence, ΔACB∼ΔPRQ (sss similarity)

∠B=∠Q

Hence, this is the answer