Question

if ∠B and ∠Q are acute angles such that sinB = sinQ, then prove that ∠B = ∠Q​

Answer 1

Step-by-step explanation:

Given sinB = sinQ 

we have sin B = AC/AB and sin Q = PR/PQ

Then AC/AB = PR/PQ

Therefore AC/AB = PR/PQ =k

By using Pythagoras theorem 

  (BC^2 =AB^2 - AC^2) / (QR^2 = PQ^2 - PR^2 )    

 (SINCE AC^2=k^2 AB^2,PR^2 = k^2 PQ^2) =AB/PQ 

HENCE, AB/PQ = AC/PR = BC/QR

therefore , B=Q

Answer 2

Question:

If ∠B and ∠Q are acute angles such that sinB = sinQ, then prove that ∠B = ∠Q.

Solution:

Using the trigonometric rules (the one given in your textbook):–

Let us consider two right triangles ABC and PQR where sinB=sinQ (refer the attachment)

As, Given that,

$sinB = \frac{opp.side}{hypotenuse} = \frac{AC }{AB}$

and, similarly,

$sinQ = \frac{PR }{PQ }$

As, sineB=sineQ

.

Now,

Let, this ratio be equal to 'x', the ;

Then,

$\frac{AC}{AB} = \frac{PR}{PQ} = x \: (let) .......(eq.n \: 1)$

So, AC=PRx .....(eq.ⁿ(a)) and,

AB=PQx.....(eq.ⁿ(b))

.

Now,

Using the Pythagoras theorem in Both the right Triangles;

In ∆ABC,

$AB {}^{2} = BC {}^{2} +AC {}^{2}$

OR

$BC = \sqrt{AB {}^{2} - AC {}^{2} }$........(2nd eq.ⁿ)

||ly, In ∆PQR,

$PQ {}^{2} =PR {}^{2} +RQ {}^{2}$

OR

$RQ = \sqrt{PQ {}^{2} - PR {}^{2} }$..........(3rd eq.ⁿ)

.

On dividing eq.ⁿ (2) by (3) gives;

$\frac{BC }{QR } = \frac{ \sqrt{AB {}^{2} - AC {}^{2} } }{ \sqrt{ PQ {}^{2} - PR {}^{2} } }$

.

[From eq.ⁿ (a) and (b)]

$\frac{BC }{QR } = \frac{ \sqrt{(xPQ) {}^{2} - (x PR) {}^{2} } }{ \sqrt{{ PQ {}^{2} - PR {}^{2} } } }$

$\frac{BC }{QR } = \frac{x \sqrt{PQ {}^{2} - PR {}^{2} } }{ \sqrt{ PQ {}^{2} - PR {}^{2} } }$

$\frac{BC}{QR } = x$

BUT, From eq.ⁿ(1);

$\frac{AC}{AB} = \frac{PR}{PQ} = x \:$

.

So,

$\frac{AC}{AB} = \frac{PR}{PQ} = \frac{BC}{QR }$

As, the ratio of corresponding sides of the two triangles ∆ABC and ∆PQR are equal. (Proved Above)

Therefore, ∆ABC~∆PQR (by SSS~ty)

As, the Corresponding parts of two similar triangles are equal.

Result:

• ∴ ∠B=∠Q (By C.P.S.T.)