Question

 If A, B and C are interior angles of a triangle ABC then cos(B+C)/2is equal to​

Answer 1

Step-by-step explanation:

Given △ABC

Given △ABCWe know that sum of three angles of a triangle is 180

Given △ABCWe know that sum of three angles of a triangle is 180Hence ∠A+∠B+∠C=180o

Given △ABCWe know that sum of three angles of a triangle is 180Hence ∠A+∠B+∠C=180oor A+B+C=180o

Given △ABCWe know that sum of three angles of a triangle is 180Hence ∠A+∠B+∠C=180oor A+B+C=180oB+C=180o−A

Given △ABCWe know that sum of three angles of a triangle is 180Hence ∠A+∠B+∠C=180oor A+B+C=180oB+C=180o−AMultiply both sides by 21

Given △ABCWe know that sum of three angles of a triangle is 180Hence ∠A+∠B+∠C=180oor A+B+C=180oB+C=180o−AMultiply both sides by 2121(B+C)=21(180o−A)

Given △ABCWe know that sum of three angles of a triangle is 180Hence ∠A+∠B+∠C=180oor A+B+C=180oB+C=180o−AMultiply both sides by 2121(B+C)=21(180o−A)21(B+C)=90o−2A...(1)

Given △ABCWe know that sum of three angles of a triangle is 180Hence ∠A+∠B+∠C=180oor A+B+C=180oB+C=180o−AMultiply both sides by 2121(B+C)=21(180o−A)21(B+C)=90o−2A...(1)Now 

Given △ABCWe know that sum of three angles of a triangle is 180Hence ∠A+∠B+∠C=180oor A+B+C=180oB+C=180o−AMultiply both sides by 2121(B+C)=21(180o−A)21(B+C)=90o−2A...(1)Now 21(B+C)

Given △ABCWe know that sum of three angles of a triangle is 180Hence ∠A+∠B+∠C=180oor A+B+C=180oB+C=180o−AMultiply both sides by 2121(B+C)=21(180o−A)21(B+C)=90o−2A...(1)Now 21(B+C)Taking sine of this angle

Given △ABCWe know that sum of three angles of a triangle is 180Hence ∠A+∠B+∠C=180oor A+B+C=180oB+C=180o−AMultiply both sides by 2121(B+C)=21(180o−A)21(B+C)=90o−2A...(1)Now 21(B+C)Taking sine of this anglesin(2B+C)[2B+C=90o−2A]

Given △ABCWe know that sum of three angles of a triangle is 180Hence ∠A+∠B+∠C=180oor A+B+C=180oB+C=180o−AMultiply both sides by 2121(B+C)=21(180o−A)21(B+C)=90o−2A...(1)Now 21(B+C)Taking sine of this anglesin(2B+C)[2B+C=90o−2A]sin(90o−2A)

Given △ABCWe know that sum of three angles of a triangle is 180Hence ∠A+∠B+∠C=180oor A+B+C=180oB+C=180o−AMultiply both sides by 2121(B+C)=21(180o−A)21(B+C)=90o−2A...(1)Now 21(B+C)Taking sine of this anglesin(2B+C)[2B+C=90o−2A]sin(90o−2A)cos2A[sin(90o−θ)=cosθ]

Given △ABCWe know that sum of three angles of a triangle is 180Hence ∠A+∠B+∠C=180oor A+B+C=180oB+C=180o−AMultiply both sides by 2121(B+C)=21(180o−A)21(B+C)=90o−2A...(1)Now 21(B+C)Taking sine of this anglesin(2B+C)[2B+C=90o−2A]sin(90o−2A)cos2A[sin(90o−θ)=cosθ]Hence sin(2

Hope it was helpful fo you.

Answer 2

Answer:

SinA/2

Step-by-step explanation:

A+B+C=180

A=180-(B+C).........(1)

Cos(B+C)/2=Sin{90-(B+C)/2}

=Sin{180-(B+C)}/2

=SinA/2 (from 1)