For a triangle ABC- show that sin B - C -2-cos A -2- where A- B and C are interior angles of ABC-

We know that ∠A + ∠B + ∠C = 180^∘ Thus we have , B + C = 180^∘ A or B + C/2 = 90^∘ A/2 or sin(B + C/2) = sin ( 90 A/2) or sin(B + C/2) = cos ( A/2) ...

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Standard X

Mathematics

Trigonometric Ratios of Standard Angles

For a triangl...

Question

For a triangle ABC, show that sin $(\frac{B + C}{2})$ = cos $(\frac{A}{2})$ , where A, B and C are interior angles of $△$ ABC.

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Solution

We know that $∠ A$ + $∠ B$ + $∠ C$ = $180^{\circ}$

Thus we have , B + C = $180^{\circ}$ - A

or $\frac{B + C}{2}$ = $90^{\circ}$ - $\frac{A}{2}$

or sin $(\frac{B + C}{2})$ = sin $(90 - \frac{A}{2})$

or sin $(\frac{B + C}{2})$ = cos $(\frac{A}{2})$

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Similar questions

Q. If

A, B

and

C

are interior angles of a triangle

A B C

, then show that

sin (\frac{B + C}{2}) = cos \frac{A}{2} .

Q. if

A, B

and

C

are interior angles of a triangle

A B C

then show that

sin (\frac{B + C}{2}) = cos \frac{A}{2}

Q. Question 6
If A, B and C are interior angles of a triangle ABC, then show that

s i n \frac{(B + C)}{2} = c o s \frac{A}{2}

Q. If A, B, C are the interior angles of a

Δ

ABC, show that

sin \frac{B + C}{2} = cos \frac{A}{2}

Say true or false:

A, B

and

C

are interior angles of a triangle

A B C

, then

sin (\frac{B + C}{2}) = cos \frac{A}{2}

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For a triangle ABC, show that sin(B+C2) = cos (A2) , where A, B and C are interior angles of △ ABC.

We know that ∠A + ∠B + ∠C = 180∘ Thus we have , B + C = 180∘ - A or B+C2 = 90∘ - A2 or sin(B+C2) = sin (90−A2) or sin(B+C2) = cos (A2)

For a triangle ABC, show that sin $(\frac{B + C}{2})$ = cos $(\frac{A}{2})$ , where A, B and C are interior angles of $△$ ABC.

We know that $∠ A$ + $∠ B$ + $∠ C$ = $180^{\circ}$

Thus we have , B + C = $180^{\circ}$ - A

or $\frac{B + C}{2}$ = $90^{\circ}$ - $\frac{A}{2}$

or sin $(\frac{B + C}{2})$ = sin $(90 - \frac{A}{2})$

or sin $(\frac{B + C}{2})$ = cos $(\frac{A}{2})$