In triangle ABC,a=6 text{ cm},b=10 text{ cm} | vedclass

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(D) Given: $a=6 	ext{ cm}$,$b=10 	ext{ cm}$,$c=14 	ext{ cm}$.Using the Law of Cosines:$\cos C = \frac{a^2 + b^2 - c^2}{2ab} = \frac{6^2 + 10^2 - 14

Vedclass · Accepted Answer

$60$

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(D) Given: $a=6 	ext{ cm}$,$b=10 	ext{ cm}$,$c=14 	ext{ cm}$.Using the Law of Cosines:$\cos C = \frac{a^2 + b^2 - c^2}{2ab} = \frac{6^2 + 10^2 - 14^2}{2 	imes 6 	imes 10} = \frac{36 + 100 - 196}{120} = \frac{-60}{120} = -\frac{1}{2}$.Since $\cos C = -\frac{1}{2}$,$C = 120^{\circ}$,which is an obtuse angle.The sum of all angles in a triangle is $180^{\circ}$.Therefore,the sum of the remaining two angles (which must be acute) is $A + B = 180^{\circ} - 120^{\circ} = 60^{\circ}$.

In $\triangle ABC$,$a=6 \text{ cm}$,$b=10 \text{ cm}$ and $c=14 \text{ cm}$. Then,the sum of the acute angles of the triangle is (in $^{\circ}$)

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