Embibe Experts Solutions for Chapter: Solution of Triangle, Exercise 2: Exercise

The sides of triangle $ABC$ are in $\mathrm{A}.\mathrm{P}.$ (order being $a,b,c$) and satisfy $\frac{2!}{1!9!}+\frac{2!}{3!7!}+\frac{1}{5!5!}=\frac{8^a}{\left(2b\right)!}$, then the value of $\cos A+\cos B$ is

If $a, b$ and $c$ are the sides of a triangle, then the minimum value of $\frac{2a}{b+c-a}+\frac{2b}{c+a-b}+\frac{2c}{a+b-c}$ is

In triangle $ABC, \angle C=\frac{2\mathrm{\pi }}{3}$ and $CD$ is the internal angle bisector of $\angle C,$ meeting the side $AB$ at $D$. Length of $CD$ is equal to

In the given figure, $AB$ is the diameter of the circle, centred at $\text{'}O\text{'}.$ If $\angle COA=60°, AB=2r, AC=d$ and $CD=l,$ then $l$ is equal to

The radii $r_1, r_2, r_3$ of the escribed circles of the triangle $ABC$ are in $\mathrm{H}.\mathrm{P}.$ If the area of the triangle is $24 {\mathrm{cm}}^2$ and its perimeter is $24 \mathrm{cm}$, then the length of its largest side is

If $\text{'}O\text{'}$ is the circumcentre of $△ABC$ and $R_1, R_2$ and $R_3$ are the radii of the circumcircles of triangles $OBC, OCA$ and $OAB,$ respectively, then $\frac{a}{R_1}+\frac{b}{R_2}+\frac{c}{R_3}$ has the value equal to

In triangle $ABC, \angle A=60°, \angle B=40°$ and $\angle C=80°$. If $P$ is the centre of the circumcircle of triangle $ABC$ with radius unity, then the radius of the circumcircle of triangle $BPC$ is

Let area of triangle $ABC$ is $\frac{\sqrt{3}-1}{2}$ sq. units and $c=\left(\sqrt{3}-1\right), b=2$ and $\angle A$ is acute. The measure of the angle $C$ is