In a triangle ABC,with usual notations a=2, b | vedclass

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

Question

(A) Using the cosine rule,$\cos A = \frac{b^2+c^2-a^2}{2bc}$,$\cos B = \frac{c^2+a^2-b^2}{2ca}$,and $\cos C = \frac{a^2+b^2-c^2}{2ab}$.Substituting th

Vedclass · Accepted Answer

$\frac{19}{30}$

Vedclass · Answer

(A) Using the cosine rule,$\cos A = \frac{b^2+c^2-a^2}{2bc}$,$\cos B = \frac{c^2+a^2-b^2}{2ca}$,and $\cos C = \frac{a^2+b^2-c^2}{2ab}$.Substituting these into the expression:$\frac{\cos A}{a}+\frac{\cos B}{b}+\frac{\cos C}{c} = \frac{b^2+c^2-a^2}{2abc} + \frac{c^2+a^2-b^2}{2abc} + \frac{a^2+b^2-c^2}{2abc}$$= \frac{b^2+c^2-a^2+c^2+a^2-b^2+a^2+b^2-c^2}{2abc}$$= \frac{a^2+b^2+c^2}{2abc}$Given $a=2, b=3, c=5$,note that $a+b=c$ $(2+3=5)$,which means the triangle is degenerate (a straight line).However,applying the formula:$= \frac{2^2+3^2+5^2}{2(2)(3)(5)} = \frac{4+9+25}{60} = \frac{38}{60} = \frac{19}{30}$.

In a triangle $ABC$,with usual notations $a=2, b=3, c=5$,then $\frac{\cos A}{a}+\frac{\cos B}{b}+\frac{\cos C}{c}=$

Similar Questions

In a $\Delta ABC,$ if $A:B:C = 3:5:4,$ then $a + b + c\sqrt{2}$ is equal to

In a $\triangle ABC$,it is known that $AB=AC$. Suppose $D$ is the mid-point of $AC$ and $BD=BC=2$. Then,the area of the $\triangle ABC$ is

The smallest angle of the triangle whose sides are $6 + \sqrt{12}$,$\sqrt{48}$,and $\sqrt{24}$ is:

In $\triangle ABC$, if $a = 13, b = 14, c = 15$, then the value of $\sin A + \cos A$ is...

With usual notations,the perimeter of a triangle $ABC$ is $6$ times the arithmetic mean of the sines of its angles. If $a=1$,then $\angle A=$

Vedclass Test Series

Exam Paper Generator

Online Exam Module