Solved Example:
If ∠ B and ∠ Q are acute angles such that sin B = sin Q, then
prove that ∠ B = ∠ Q.
A
C
B
P
R
Q
Proof:
Consider ΔACB and ΔPRQ in which ∠C = ∠R = 90º,
∠B and ∠Q are acute.
sin B = sin Q
AC
AB
sin B =
PR
PQ
sin Q =
=
=
AC
AB
PR
PQ
AC
PR
AB
PQ
∴
=
k
...(i)
AC
=
=
AC
PR
AB
PQ
k PR,
AB
=
k PQ
∴
...(ii)
Let
AC
=
k PR,
AB
=
k PQ
...(ii)
(kPQ)2
=
(kPR)²
+
BC²
AB²
=
AC²
+
BC²
[Pythagoras theorem]
=
k2
BC2
∴
∴
...(iii)
PQ²
=
PR²
+
QR²
[Pythagoras theorem]
Now,
BC2
QR2
=
k2
=
k …(i)
=
AC
PR
AB
PQ
Solved Example:
A
C
B
P
R
Q
Proof:
QR²
=
PQ²
–
PR²
∴
...(iv)
If ∠ B and ∠ Q are acute angles such that sin B = sin Q, then
prove that ∠ B = ∠ Q.
(PQ2
–
PR2)
(PQ2
–
PR)2
(PQ2
–
PR)2
∴
BC2
QR2
=
k2
BC2
=
k²
–
k²
∴
PQ2
PR²
[Dividing (iii) and (iv)]
=
k …(i)
=
AC
PR
AB
PQ
Solved Example:
A
C
B
P
R
Q
Proof:
If ∠ B and ∠ Q are acute angles such that sin B = sin Q, then
prove that ∠ B = ∠ Q.
∴
BC2
QR2
=
k2
∴
BC
QR
=
k
…(v)
In ΔACB and ΔPRQ,
AC
PR
=
AB
PQ
=
BC
QR
…[From (i) and (v)]
∴
ΔACB ~ ΔPRQ
(By SSS similarity criterion)
∴
∠B
=
∠Q
…(Corresponding angles of similar triangles)
(Taking square roots)