In triangle ABC- with usual notation- then cosA-b2-c2-a22bc- cosB-c2-a2-b22ca-cosC-a2-b2-c22ab

The correct option is A TrueAccording to the cosine rule in any triangle #160; a^2=b^2+c^2 2bccosA So #160; Cos A=b^2+c^2 a^22bc Similarly Cos ...

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

Equal Angles Subtend Equal Sides

In triangle ...

Question

In triangle $△ A B C$ , with usual notation, then
$cos A = \frac{b^{2} + c^{2} - a^{2}}{2 b c}$ , $cos B = \frac{c^{2} + a^{2} - b^{2}}{2 c a}, cos C = \frac{a^{2} + b^{2} - c^{2}}{2 a b}$

True

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

False

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

Open in App

Solution

Suggest Corrections

Similar questions

△ A B C, a : b : c = cos A : cos B : cos C

△ A B C

cos A = \frac{b^{2} + c^{2} - a^{2}}{2 b c}

cos B = \frac{c^{2} + a^{2} - b^{2}}{2 c a}

cos C = \frac{a^{2} + b^{2} - c^{2}}{2 a b}

Δ A B C

c o s A = \frac{b^{2} + c^{2} - a^{2}}{2 b c}, c o s B = \frac{c^{2} + a^{2} - b^{2}}{2 a c}, c o s C = \frac{a^{2} + b^{2} - c^{2}}{2 a b}

A B C

\frac{b^{2} - c^{2}}{cos B + cos C} + \frac{c^{2} - a^{2}}{cos C + cos A} + \frac{a^{2} - b^{2}}{cos A + cos B} = 0.

\frac{a^{2} - b^{2}}{cos A + cos B} + \frac{b^{2} - c^{2}}{cos B + cos C} + \frac{c^{2} - a^{2}}{cos C + cos A} = 0

Join BYJU'S Learning Program