in triangle ABC, prove that cos (a+b)+cosc=0
Answers
Answered by
0
Step-by-step explanation:
ho thank u for giving the question
Answered by
1
Answer:
In ΔABC, Let
BC
=
a
,
CA
=
b
and
AB
=
c
, then
a=∣
BC
∣,b=∣
CA
∣ and c=∣
AB
∣
From ΔABC, we get
⇒
BC
+
CA
=
BA
⇒
BC
+
CA
+
AB
=0
⇒
a
+
b
+
c
=0
⇒
a
×(
a
+
b
+
c
)=
a
×
0
⇒
a
×
a
+
a
×
b
+
a
×
c
=0
⇒
a
×
b
=−
a
×
c
⇒
a
×
b
=
c
×
a
...... (i)
Similarly,
a
×
b
=
b
×
c
...... (ii)
From (i) and (ii), we get
a
×
b
=
b
×
c
=
c
×
a
∣
a
×
b
∣=∣
b
×
c
∣=∣
c
×
a
∣
absin(π−C)=bcsin(π−A)=casin(π−B)
absinC=bcsinA=casinB
Dividing by abc throughout, we get
c
sinC
=
a
sinA
=
b
sinB
⇒
sinA
a
=
sinB
b
=
sinC
c
Similar questions