Question

In a $△ABC$ , i f b=20 , c=21 and $sinA=\frac{3}{5}$ then find a.

Answer 1

Given that,

$b=20$

$c=21$

$\sin A=\frac{3}{5}$

$\cos A=\sqrt{1-{\sin }^{2}A}$

$=\sqrt{1-{\left(\frac{3}{5}\right)}^2}$

$=\sqrt{1-\frac{9}{25}}$

$=\sqrt{\frac{25-9}{25}}$

$=\sqrt{\frac{16}{25}}$

$=\frac{4}{5}$

$\therefore \cos A=\frac{4}{5}$

We know that,

$\cos A=\frac{b^2+C^2-A^2}{2bc}$

$\Rightarrow \frac{4}{5}=\frac{{20}^2+{21}^2-a^2}{2\times 20\times 21}$

$\frac{4}{5}=\frac{400+441-a^2}{8\times 21}$

$4\times 8\times 21=841-a^2$

$\Rightarrow 672-841=-a^2$

$\Rightarrow a^2=49$

$\Rightarrow a=\sqrt{49}$

$\Rightarrow a=7$

Hence, $a=7$