if for triangle ABC , COSACOSB+SINASINBSINC = 1 . THEN SOLVE THE TRIANGLE
Answers
Multiplying both sides by 2 in given equality cosAcosB+sinAsinBsinC=1, we get
2cosAcosB+2sinAsinBsinC=2 or
2cosAcosB+2sinAsinBsinC=(sin2A+cos2A)+(sin2B+cos2B)
or (cos2A+cos2B−2cosAcosB)+(sin2A+sin2B−2sinAsinB)+2sinAsinB−2sinAsinBsinC=0or
or (cosA−cosB)2+(sinA−sinB)2+2sinAsinB(1−sinC)=0
Note that all three terms contained above are positive, as while first two terms are squares and hence positive, third term is positive as sine of angles A and B is positive (as they are less than π being angles of a triangle) and (1−sinC) too will be positive as sinC<1.
But, their sum is zero and hence each term is equal to zero, i.e.
cosA−cosB=0, sinA−sinB=0 and 1−sinC=0 or
cosA=cosB, sinA=sinB and 1−sinC=0
i.e. A=B and sinC=1 i.e. C=π2
Hence the triangle is isosceles and right angled.
Step-by-step explanation:
Multiplying both sides by
2
in given equality
cos
A
cos
B
+
sin
A
sin
B
sin
C
=
1
, we get
2
cos
A
cos
B
+
2
sin
A
sin
B
sin
C
=
2
or
2
cos
A
cos
B
+
2
sin
A
sin
B
sin
C
=
(
sin
2
A
+
cos
2
A
)
+
(
sin
2
B
+
cos
2
B
)
or
(
cos
2
A
+
cos
2
B
−
2
cos
A
cos
B
)
+
(
sin
2
A
+
sin
2
B
−
2
sin
A
sin
B
)
+
2
sin
A
sin
B
−
2
sin
A
sin
B
sin
C
=
0
or
or
(
cos
A
−
cos
B
)
2
+
(
sin
A
−
sin
B
)
2
+
2
sin
A
sin
B
(
1
−
sin
C
)
=
0
Note that all three terms contained above are positive, as while first two terms are squares and hence positive, third term is positive as sine of angles
A
and
B
is positive (as they are less than
π
being angles of a triangle) and
(
1
−
sin
C
)
too will be positive as
sin
C
<
1
.
But, their sum is zero and hence each term is equal to zero, i.e.
cos
A
−
cos
B
=
0
,
sin
A
−
sin
B
=
0
and
1
−
sin
C
=
0
or
cos
A
=
cos
B
,
sin
A
=
sin
B
and
1
−
sin
C
=
0
i.e.
A
=
B
and
sin
C
=
1
i.e.
C
=
π
2
Hence the triangle is isosceles and right angled.