Question

If A, B and C are interior angles of a triangle ABC, then show that sin [(A+ C)/7] = cos A/3.​

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Answer 1

Answer:

$\sin( \frac{b + c}{2} ) = \cos( \frac{a}{2} )$

Step-by-step explanation:

given: ∆ABC

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

Hence ZA + B + 4C = 180°

or A+B+C = 180⁰

B+C = 180° - A

$Multiply both sides by \frac{1}{2 } \\ \frac{1}{2} (B+C) = \frac{1}{2} (180° − a) \\ \frac{1}{2} (b + c) = 90 {}^{0} - \frac{a}{2} ...(1) \\ now \\ \frac{1}{2} (b + c ) \\ talking \: sine \: of \: this \: angle \: \\ \sin( \frac{b + c}{2} ) ( \frac{b + c}{2} = 90 {}^{0} - \frac{a}{2} \\ \sin(90 {}^{0} - \frac{a}{2} ) \\ \cos( \frac{a}{2} ) ( \sin(90 {}^{0} - 360 {}^{0} ) = \cos( 360{}^{0} ) \\ hence \: \sin( \frac{b + c}{2} ) = \cos( \frac{a}{2} ) proved$

Answer 2

Answer:

hlo g

gud morning

aap baat kyu ni krteʕಠ_ಠʔ

Step-by-step explanation:

koina choro

tc