Question

In a ∆ABC, if cos a = SinB/2SinC show that ∆ is isocles​

Answer 1

$<b> Heya! </b>$

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$<u>★ Trigonometry★ </u>$

Given that ,

$= > \: cos \: a \: = \frac{sin \: b}{2 \: sin \: c} \\ \\ = > let \: us \: assume \: the \: following \: \\ \\ = \frac{sin \: a}{a} = \frac{sin \: b}{b} = \frac{sin \: c \: }{c} = k$

★ From this assumption we get ,

$<b> Sin a = k• a </b>$

$<b> Sin b = k•b</b>$

$<b> Sin c = k•c </b>$

Now , from the given relation :

$= > cos \: a \: = \frac{sin \: b}{2 \: sin \: c} \\ \\ = > 2 \: cos \: a \times \: sin \: c \: = sin \: b \\ \\ = > 2( \frac{b {}^{2} + c {}^{2} - a {}^{2} }{2bc} ) \times k.c = k.b$

This gives us →

=> b² + c² - a² = b²

→ $<b> c² = a²</b>$

°•° Two sides have been proved equal .Hence the ∆ is an isoceles triangle !!


$<i> <u> Hence Proved !! </u> </i>$
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Answer 2

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