0.1,
If √B and √ Q are acute angles such
that sin b=sinQ
then prove that √b=√Q
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Answer:
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
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