if angle B and angle Q are acute angle of two different triangles such that sinB=SinQ, then prove that angle B= angle Q
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Answer:
there are the two steps of this sum,
First step :
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
now,nextstep
second step :
sinB−sinQ=0⇒2cos(
2
B+Q
)sin(
2
B−Q
)=0
⇒cos(
2
B+Q
)=0 or sin(
2
B−Q
)=0
⇒
2
B+Q
=
2
π
2
B−Q
=0⇒
∠B=∠Q
⇒B+Q=π But A+B+C=π⇒A=0 Not possible
Step-by-step explanation:
hope it helps you...
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