In a triangle, if $b=20, c=21$ and $\sin A=\frac{3}{5}$, then $a$ is equal to :

Question

Aicharya · Accepted Answer

We have, $b=20, c=21$ and $\sin A=\frac{3}{5}$ Now, $\cos ^2 A=1-\sin ^2 A=1-\left(\frac{3}{5}ight)^2$ $=1-\frac{9}{25}=\frac{16}{25}$ $\Rightarrow \quad \cos A=\frac{4}{5}$ Now, $\cos A=\frac{b^2+c^2-a^2}{2 b c}$ $\Rightarrow \quad \frac{4}{5}=\frac{(20)^2+(21)^2-a^2}{2 \cdot 20 \cdot 21}$ $\Rightarrow \quad 400+441-a^2=672$ $\Rightarrow \quad a^2=841-672=169$ $	herefore \quad a=13$

In a triangle, if $b=20, c=21$ and $\sin A=\frac{3}{5}$, then $a$ is equal to :

Solution