EXERCISE 8.1
Q.6) If ∠A and ∠B are acute angles such that cos A = cos B, then
prove that ∠A = ∠B.
Proof:
Consider ΔAMN and ΔBPQ in which ∠M = ∠P = 90º,
∠A and ∠B are acute.
cos A = cos B
Let
A
M
N
B
P
Q
AM
AN
cos A =
BP
BQ
cos B =
=
=
AM
AN
BP
BQ
AM
BP
AN
BQ
∴
=
k
...(i)
AM
=
=
AM
BP
AN
BQ
k BP,
AN
=
k BQ
∴
...(ii)
EXERCISE 8.1
Q.6) If ∠A and ∠B are acute angles such that cos A = cos B, then
prove that ∠A = ∠B.
Proof:
A
M
N
B
P
Q
AM
=
k BP,
AN
=
k BQ
...(ii)
(kBQ)²
=
(kBP)²
+
MN²
k2BQ²
=
k2BP²
MN²
AN²
=
AM²
+
MN²
[Pythagoras theorem]
–
(BQ²
=
BP²
MN²
–
k2
)
∴
∴
∴
...(iii)
BQ²
=
BP²
+
PQ²
[Pythagoras theorem]
BQ²
=
BP²
PQ²
–
∴
...(iv)
Now,
MN2
PQ2
=
k2
(BQ²
BP²)
–
(BQ²
BP²)
–
[Dividing (iii) and (iv)]
=
k …(i)
=
AM
BP
AN
BQ
EXERCISE 8.1
Q.6) If ∠A and ∠B are acute angles such that cos A = cos B, then
prove that ∠A = ∠B.
Proof:
A
M
N
B
P
Q
Now,
MN2
PQ2
=
k2
(BQ²
BP²)
–
(BQ²
BP²)
–
=
k2
MN2
PQ2
MN
PQ
k
...(v)
=
∴
=
AM
BP
AN
BQ
MN
PQ
=
∴
Δ AMN
~
Δ BPQ
[By SSS similarity criterion]
∴
∠A
=
∠B
[corresponding angles of
similar triangles]
∴
[Taking square roots]
=
k …(i)
=
AM
BP
AN
BQ
In ΔAMN and ΔBPQ,
[From (i) and (v)]