Properties Of Triangles - Trigonometry - Mathematics Previous Year Questions

In $\triangle A B C$, if $A$ is acute, $C$ is obtuse, $\sin A=\frac{3 \sqrt{3}}{14}, a=3$ and $b=5$, then $c=$

If $\Delta$ denotes the area of $\triangle A B C$, then $(b \sin C+c \sin B)(b \cos C+c \cos B)=$

Let $A$ be the area of in-circle and $A_1, A_2, A_3$ be the area of ex-circles of a triangle. If $A_1=4, A_2=9, A_3=16$, then $A=$

In a $\triangle A B C$, if $(b+c)^2 \sin ^2 \frac{A}{2}+(b-c)^2 \cos ^2 \frac{A}{2}=K(1-\cos 2 A)$, then $K=$

In $\triangle A B C$, if $B C$ is the hypotenuse, then $r_2+r_3=$

In a $\triangle A B C$, if $b=7, c=4 \sqrt{3}$ and $A=\frac{\pi}{6}$ then a $\sin B \sin C=$

In $\triangle A B C$, if $\frac{\cos A}{a}=\frac{\cos B}{b}=\frac{\cos \cdot C}{c}$ and side $a=2$, then area of the $\triangle A B C$ (in sq units) is

In $\triangle A B C$, if $b=6, c=7$ and $\tan \frac{A}{2}=\frac{1}{\sqrt{6}}$, then the inradius of $\triangle A B C$ is

In $\triangle A B C$, if $a=7, b=8$ and $c=9$, then $\frac{1}{r_1^2}+\frac{1}{r_2^2}+\frac{1}{r_3^2}=$

In $\triangle A B C$, if $A$ is an acute angle, $b=6, c=9$ and $\sin A=\frac{2 \sqrt{14}}{9}$, then $3 a(\cos B+\cos C)=$

If the roots of the equation $x^3-11 x^2+36 x-36=0$ are the ex-radii of a $\triangle A B C$, then the perimeter of the $\triangle A B C$ is

In $\triangle A B C$, if $a, b, c$ are in arithmetic progression and $A=2 C$, then $b: c=$

Assertion (A) In $\triangle A B C$, if $r=6, r_2=36, R=15$, then $c^2+a^2=b^2$.

Reason (R) In $\triangle A B C$, if $r: R: r_2=1: 2.5: 6$, then $B=90^{\circ}$. The correct option among the following is

In $\triangle A B C$, if $a: b: c=4: 5: 6$, then the ratio of the circumradius to its inradius is

The perimeter of a $\triangle A B C$ is 6 times the arithmetic mean of the values of the sine of its angles. If its side $B C$ is of unit length, then $\angle A=$

In a $\triangle A B C, \frac{\Delta^2}{a^2+b^2+c^2}\left(\frac{1}{r_1^2}+\frac{1}{r_2^2}+\frac{1}{r_3^2}+\frac{1}{r^2}\right)=$

In a $\triangle A B C,\left(b^2-c^2\right) \cot A+\left(c^2-a^2\right) \cot B=$

If $R: r_1: r=5: 12: 2$, then $r+r_3+r_2-r_1=$

In $\triangle A B C, \frac{1+\cos C}{r_1+r_2}+\frac{1+\cos A}{r_2+r_3}+\frac{1+\cos B}{r_1+r_3}=$