In a triangle ABC, if sin A sin B = frac{ab}{ | vedclass

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

Question

(C) Given the relation: $\sin A \sin B = \frac{ab}{c^2}$ Using the Sine Rule, we know that $\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C} = 2

Vedclass · Accepted Answer

right angled

Vedclass · Answer

(C) Given the relation: $\sin A \sin B = \frac{ab}{c^2}$ Using the Sine Rule, we know that $\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C} = 2R$. This implies $\sin A = \frac{a}{2R}$, $\sin B = \frac{b}{2R}$, and $\sin C = \frac{c}{2R}$. Substituting these into the given equation: $\left(\frac{a}{2R}ight) \left(\frac{b}{2R}ight) = \frac{ab}{c^2}$ $\frac{ab}{4R^2} = \frac{ab}{c^2}$ Since $a, b 
eq 0$, we can cancel $ab$ from both sides: $\frac{1}{4R^2} = \frac{1}{c^2}$ $\Rightarrow c^2 = 4R^2$ $\Rightarrow c = 2R$. Since $c = 2R$, we have $\frac{c}{\sin C} = 2R \Rightarrow \sin C = \frac{c}{2R} = \frac{2R}{2R} = 1$. Therefore, $C = 90^{\circ}$. Thus, the triangle is a right-angled triangle.

In a triangle $ABC$, if $\sin A \sin B = \frac{ab}{c^2}$, then the triangle is

Similar Questions

In a $\triangle ABC$, $a^2 \sin 2C + c^2 \sin 2A$ is equal to (in $\Delta$)

If $p_1, p_2, p_3$ are altitudes of a triangle $ABC$ from the vertices $A, B, C$ respectively and if $\Delta$ is the area of the triangle,$S$ is the semi-perimeter of the triangle,then find the value of $\frac{\cos A}{p_1} + \frac{\cos B}{p_2} + \frac{\cos C}{p_3}$.

In a $\triangle ABC$,$2ac \sin \frac{1}{2}(A-B+C)$ is equal to

In a triangle $ABC$ with usual notations $a=2$ and $b=3$,the value of $\frac{\cos 2A}{a^2} - \frac{\cos 2B}{b^2}$ is:

In $\triangle ABC$,if $A=75^{\circ}$ and $B=45^{\circ}$,then the value of $b+c\sqrt{2}$ is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module