Question

if a,b,c are interior angles of triangle abc ,then show that sin a+b/2=cos c/2
​

Answer 1

Step-by-step explanation:

Given △ABC

We know that sum of three angles of a triangle is 180

Hence ∠A+∠B+∠C=180

o

or A+B+C=180

o

B+C=180

o

−A

Multiply both sides by

2

1

2

1

(B+C)=

2

1

(180

o

−A)

2

1

(B+C)=90

o

−

2

A

...(1)

Now

2

1

(B+C)

Taking sine of this angle

sin(

2

B+C

)[

2

B+C

=90

o

−

2

A

]

sin(90

o

−

2

A

)

cos

2

A

[sin(90

o

−θ)=cosθ]

Hence sin(

2

B+C

)=cos

2

A

proved

Answer 2

Answer:

a+b/2=cos c/2

Step-by-step explanation:

sin (a+b)/2 = cos 90 - (a+b)/2

= cos [180 - (a+b)]/2

= cos [180 - a - b]/2

= cos c/2