With usual notation,in triangle ABC,m angle A | vedclass

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

Question

(B) Given the expression $E = \left(1+\frac{a}{c}+\frac{b}{c}\right)\left(1+\frac{c}{b}-\frac{a}{b}\right)$.This can be rewritten as $E = \left(\frac{

Vedclass · Accepted Answer

$2+\sqrt{3}$

Vedclass · Answer

(B) Given the expression $E = \left(1+\frac{a}{c}+\frac{b}{c}ight)\left(1+\frac{c}{b}-\frac{a}{b}ight)$.This can be rewritten as $E = \left(\frac{c+a+b}{c}ight)\left(\frac{b+c-a}{b}ight)$.$E = \frac{(b+c)+a}{c} 	imes \frac{(b+c)-a}{b} = \frac{(b+c)^2 - a^2}{bc}$.Expanding the numerator: $E = \frac{b^2 + c^2 + 2bc - a^2}{bc} = \frac{b^2 + c^2 - a^2}{bc} + 2$.Using the Law of Cosines,$a^2 = b^2 + c^2 - 2bc \cos A$,so $b^2 + c^2 - a^2 = 2bc \cos A$.Substituting this into the expression: $E = \frac{2bc \cos A}{bc} + 2 = 2 \cos A + 2$.Given $A = 30^{\circ}$,$\cos 30^{\circ} = \frac{\sqrt{3}}{2}$.Thus,$E = 2(\frac{\sqrt{3}}{2}) + 2 = \sqrt{3} + 2$.

With usual notation,in triangle $ABC$,$m \angle A = 30^{\circ}$. Then the value of $\left(1+\frac{a}{c}+\frac{b}{c}\right)\left(1+\frac{c}{b}-\frac{a}{b}\right)$ is equal to

Similar Questions

The angles of a triangle are in the ratio $3: 5: 10$. Then the ratio of the smallest side to the greatest side is:

If the sides of a triangle are $p, q$ and $\sqrt{p^2 + pq + q^2}$,then the largest angle is

The ratio of the sides of triangle $ABC$ is $1:\sqrt{3}:2$. The ratio of angles $A:B:C$ is

If $A$ is the area and $2s$ is the sum of the $3$ sides of a triangle,then :

In $\triangle ABC$,with usual notations,the value of $\frac{b \sin B - c \sin C}{\sin (B - C)}$ is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module