If the sides of a triangle are 6, 10 and 14,t | vedclass

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(A) Let the sides be $a = 6$,$b = 10$,and $c = 14$. To determine the type of triangle,we use the Law of Cosines to find the angle $	heta$ opposite to

Vedclass · Accepted Answer

Obtuse angled

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(A) Let the sides be $a = 6$,$b = 10$,and $c = 14$. To determine the type of triangle,we use the Law of Cosines to find the angle $	heta$ opposite to the longest side $c = 14$. $\cos 	heta = \frac{a^2 + b^2 - c^2}{2ab}$ $\cos 	heta = \frac{6^2 + 10^2 - 14^2}{2 	imes 6 	imes 10}$ $\cos 	heta = \frac{36 + 100 - 196}{120}$ $\cos 	heta = \frac{136 - 196}{120} = \frac{-60}{120} = -\frac{1}{2}$ Since $\cos 	heta = -\frac{1}{2}$,we have $	heta = 120^\circ$. Since one angle is greater than $90^\circ$,the triangle is an obtuse-angled triangle.

If the sides of a triangle are $6, 10$ and $14$,then the triangle is

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