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

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(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

Vedclass · Answer

(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

Similar Questions

In any triangle $ABC$,what is the value of $a(b \cos C - c \cos B)$?

In $\triangle ABC$,if $\frac{1}{a+b} + \frac{1}{c+a} = \frac{3}{a+b+c}$,then $\sin A$ is equal to

The ratios of the sides in a triangle are $5:12:13$ and its area is $270 \text{ cm}^2$. The sides of the triangle in $\text{cm}$ are

In an acute angled triangle,$\cot B \cot C + \cot A \cot C + \cot A \cot B$ is equal to

If $\cos^2 A + \cos^2 C = \sin^2 B$,then $\Delta ABC$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module