In a triangle ≤ft(A+B/2)=

Correct Answer is : sin⁡c2 c/2sin2c

Correct Answer is : sin⁡c2 c/2sin2c

Test Index

AP DSC School Assistant Math Model Test 3

Section: Subject Specific

Question : 109 of 160

Marks: +1, -0

In a triangle

\cos \left(\frac{A+B}{2}\right)=\ldots

Solution:

Concept:

The sum of the interior angles of a triangle is

180^\circ

. Using this, express

\frac{A+B}{2}

in terms of

C

and apply the complementary angle identity

\cos(90^\circ - \theta) = \sin \theta

Explanation:

In any triangle, we have

A + B + C = 180^\circ

Therefore,

A + B = 180^\circ - C

Divide both sides by

2

\frac{A+B}{2} = \frac{180^\circ - C}{2} = 90^\circ - \frac{C}{2}

Now take cosine on both sides:

\cos\left(\frac{A+B}{2}\right) = \cos\left(90^\circ - \frac{C}{2}\right)

Using the identity

\cos(90^\circ - \theta) = \sin \theta

, we get:

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

Answer:

The correct option is D:

\sin\left(\frac{C}{2}\right)

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