Question

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

Class 10

Trigonometric ratios

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Answer 1

Heya!!

Here is your solution :

Given,

In a right angled triangle <Q and <B are acute angles, so

⇒( <B + <Q + 90°) = 180°

⇒( <B + <Q) = 180° - 90°

⇒( <B + <Q) = 90°

.°. <B = ( 90° - <Q ) - - - - - - - ( 1 )

According to question,

⇒sin B = sin Q

Substitute the value of ( 1 ),

⇒sin ( 90° - Q ) = sin Q

So,

⇒( 90° - Q ) = Q

⇒90° = Q + Q

⇒90° = 2Q

⇒Q = 90° ÷ 2

.°. Q = 45° -------- ( 2 )

Substitute the value of ( 2 ) in ( 1 ),

⇒<B = 90° - Q

⇒ <B = 90° - 45°

.°. <B = 45° --------- ( 3 )

From ( 2 ) and ( 3 ) , we get,

⇒<B = <Q

Proved!!

Note : The sign '<' is used to show the sign of angle.

Answer 2

$\bf \huge{hey}$

$\huge{here \: is \: the \: answer}$

⬇✔⬇✔⬇⬇⬇⬇⬇✔⬇✔⬇✔✔⬇✔⬇✔⬇

$\bf{ \angle \: b = \angle \: q}$
we have to prove that

here we can write
$\bf{( \angle \: b + \angle \: q + 90 \degree) = 180 \degree}$

$\bf{ \angle \:b + \angle \: q = 180 \degree - 90 \degree}$

$\bf{ \angle \:b + \angle \: q = 90 \degree}$

here we can write

<b=90°-<q..........(1)


A/Q

sin(90° -q)= sinq

here

90°-q =q

90°=q+q

2q= 90°

q= 45°

we can write value of q in equation 1 ..
my

then

$\bf{ \angle \: b = 90 \degree - \angle \: q}$

angle b= 90°-45°

<b= 45°

$\bf{hence \: \angle \: b = \angle \:q}$

$\huge{ \mathfrak{hope \: it \: helps}}$

$\bf \huge \boxed{thanks}$