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Byju's Answer
Standard X
Mathematics
Construction of Incircle
Let a,b and c...
Question
Let
a
,
b
and
c
be the sides opposite to angles
A
,
B
and
C
in a triangle
A
B
C
.
If
sin
A
(
sin
A
+
cos
B
−
sin
B
)
+
cos
A
(
cos
A
+
sin
B
+
cos
B
)
=
1
+
sin
C
and
a
=
4
,
b
=
3
,
then area (in sq. units) of the triangle
A
B
C
is
Open in App
Solution
sin
A
(
sin
A
+
cos
B
−
sin
B
)
+
cos
A
(
cos
A
+
sin
B
+
cos
B
)
=
1
+
sin
C
⇒
(
sin
2
A
+
cos
2
A
)
+
(
sin
A
cos
B
+
cos
A
sin
B
)
+
(
cos
A
cos
B
−
sin
A
sin
B
)
=
1
+
sin
C
⇒
1
+
sin
(
A
+
B
)
+
cos
(
A
+
B
)
=
1
+
sin
C
⇒
1
+
sin
C
−
cos
C
=
1
+
sin
C
(
∵
A
+
B
+
C
=
π
)
⇒
cos
C
=
0
∴
C
=
π
2
∴
Δ
=
1
2
a
b
sin
C
=
1
2
×
4
×
3
=
6
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Similar questions
Q.
Assertion :If in a triangle
A
B
C
,
cos
A
cos
B
+
sin
A
sin
B
sin
C
=
1
, then it is an acute angle triangle Reason: In a triangle
A
B
C
,
sin
A
:
sin
B
:
sin
C
=
a
:
b
:
c
.
Q.
If in
△
A
B
C
,
a
cos
A
+
b
cos
B
+
c
cos
C
a
sin
B
+
b
sin
C
+
c
sin
A
=
a
+
b
+
c
9
R
,
then the
△
A
B
C
is
Q.
If in a triangle
A
B
C
,
cos
A
cos
B
+
sin
A
sin
B
sin
C
=
1
,
then the sides are proportional to
Q.
In a triangle
△
A
B
C
,
3
sin
A
+
4
cos
B
=
6
and
4
sin
B
+
3
cos
A
=
1
, then the angle
C
is
Q.
If a, b, c are sides of a
△
A
B
C
, such that A, B, C are angles opposite to sides a, b, c respectively then
Δ
=
∣
∣ ∣
∣
a
2
b
sin
A
c
sin
A
b
sin
A
1
cos
A
c
sin
A
cos
A
1
∣
∣ ∣
∣
equals to
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