Question

In triangle $ ABC $, if $ a = 13 $, $ b = 8 $, $ c = 7 $, then $ \cos(B+C) = $

• A. $ \frac{11}{13} $
• B. $ \frac{23}{26} $
• C. $ \frac{3}{4} $
• D. $ \frac{1}{2} $

Hint:

When finding $ \cos(B+C) $ in a triangle, use the angle sum property $ B + C = 180^\circ - A $ and the identity $ \cos(180^\circ - A) = -\cos A $. Then apply the Law of Cosines to compute $ \cos A $.

Answer 1

The Correct Option is D

Step 1: Use the angle sum property of a triangle.
In any triangle, the sum of angles is $ 180^\circ $, so: $$ A + B + C = 180^\circ \quad \Rightarrow \quad B + C = 180^\circ - A. $$ Thus: $$ \cos(B+C) = \cos(180^\circ - A) = -\cos A. $$ Step 2: Use the Law of Cosines to find $ \cos A $.
The Law of Cosines states: $$ \cos A = \frac{b^2 + c^2 - a^2}{2bc}. $$ Substitute $ a = 13 $, $ b = 8 $, $ c = 7 $: $$ \cos A = \frac{8^2 + 7^2 - 13^2}{2 \cdot 8 \cdot 7}. $$ Simplify: $$ \cos A = \frac{64 + 49 - 169}{112} = \frac{113 - 169}{112} = \frac{-56}{112} = -\frac{1}{2}. $$ Step 3: Compute $ \cos(B+C) $.
Since $ \cos(B+C) = -\cos A $: $$ \cos(B+C) = -\left(-\frac{1}{2}\right) = \frac{1}{2}. $$ Step 4: Final Answer.
$$ \boxed{\frac{1}{2}} $$