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

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

Question

(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

$60$

Vedclass · Answer

(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 $^{\circ}$)

Similar Questions

In any $\triangle ABC$,$a(b \cos C - c \cos B)$ equals

In a triangle $ABC$,if $\cot \frac{A}{2} \cot \frac{B}{2} = K$,then all the possible values of $K$ lie in

If in a $\triangle ABC$,$\frac{1}{a+c} + \frac{1}{b+c} = \frac{3}{a+b+c}$,then $\angle C$ is equal to (in $^{\circ}$)

In a triangle $ABC$,let $AB = \sqrt{23}$,$BC = 3$,and $CA = 4$. Then the value of $\frac{\cot A + \cot C}{\cot B}$ is:

In a $\Delta ABC$,if $C = 90^{\circ}$,then $\frac{a^{2} + b^{2}}{a^{2} - b^{2}} \sin(A - B)$ is equal to:

Vedclass Test Series

Exam Paper Generator

Online Exam Module