If cos A2-b-c2c- then -

Cos nbsp;A2=b+c2c⇒s(s a)bc=b+c2c⇒2s2s 2a=2bb+c⇒b+c+ab+c a=2b2+2bc⇒b+c2 a2=2b2+2bc⇒c2=a2+b2 ...

You visited us 1 times! Enjoying our articles? Unlock Full Access!

Trigonometric Ratios of Half Angles

If cos A2=b+c...

Question

If $cos \frac{A}{2} = \sqrt{\frac{b + c}{2 c}}$ , then :

a^{2} + b^{2} = c^{2}

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

c^{2} + b^{2} = a^{2}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

a^{2} + c^{2} = b^{2}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

Suggest Corrections

Similar questions

cos \frac{A}{2} = \sqrt{\frac{b + c}{2 c}}

\pm \sqrt{a^{2} + b^{2} + c^{2}}

\pm \sqrt{a^{2} + b^{2} - c^{2}}

\pm \sqrt{c^{2} - a^{2} - b^{2}}

If $\frac{c + i}{c - i}$ = a+ib, where a,b,c are real, then $a^{2} + b^{2}$ =

Find:
$(a + b + c) \times (a + b - c) =$

\frac{a}{b}

\frac{(a sinθ - b cosθ)}{(a \sin θ + b cosθ)}

\frac{(a^{2} + b^{2})}{(a^{2} - b^{2})}

\frac{(a^{2} - b^{2})}{(a^{2} + b^{2})}

\frac{a^{2}}{(a^{2} + b^{2})}

\frac{b^{2}}{(a^{2} + b^{2})}

Join BYJU'S Learning Program