Question

Abc is triangle if sin (a+b/ 2) =3/2, then the value of sin c/2 is

Answer 1

we know that the sum of angle = 180

A+ B + C = 180

A+B= 180-C

$\sin( \frac{a + b}{2} ) = \frac{3}{2} \\ \sin( \frac{180 - c}{2} ) = \frac{3}{2} \\ \sin( \frac{180}{2} - \frac{c}{2} ) = \frac{3}{2} \\ \sin(90 - \frac{c}{2} ) = \frac{3}{2} \\ \cos( \frac{c}{2} ) = \frac{3}{2} = \frac{b}{h}$

second

${ \sin }^{2} (\frac{c}{2}) + { \cos }^{2} (\frac{c}{2} )= 1 \\ { \sin }^{2} (\frac{c}{2} ) + { (\frac{3}{2}) }^{2} = 1 \\ { \sin }^{2} \frac{c}{2} = 1 - \frac{9}{4} \\ = \frac{4 - 9}{4 } \\ = \frac{ - 5}{4}$

answer

Answer 2

$\sin\left (\frac{C}{2} \right )= \sqrt{\frac{5}{4}}$

Step-by-step explanation:

if ABC is a triangle

then sum of all the angles of the triangle = 180 degree

(by angle sum property )

i.e

$\angle A +\angle B+ \angle C = 180 ^\circ$

then  $\angle A + \angle B = 180 - \angle C$

Now,

$\sin \left ( \frac{a+b}{2} \right ) = \frac{3}{2}\\\\\sin \left ( \frac{180-C}{2} \right )= \frac{3}{2}\\\\\sin\left ( \frac{180}{2}- \frac{C}{2} \right )= \frac{3}{2}\\\\\sin\left ( 90- \frac{C}{2} \right )= \frac{3}{2}\\\\sin (90-\theta )= \cos\theta \\\\\cos (\frac{C}{2})= \frac{3}{2}\\\\then \\\\\sin^2\theta + \cos^2\theta =1 \\\\\sin^2\left (\frac{C}{2} \right )+ \cos^2\left ( \frac{C}{2} \right )= 1\\\\\sin^2\left (\frac{C}{2} \right )= 1- \left ( \frac{3}{2} \right )^2$

$\sin\left (\frac{C}{2} \right )= \sqrt{\frac{5}{4}}$

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