Question

in a triangle if b=20 c= 21 and sinA=3/5 find a ?

Answer 1

Given:

• In a triangle $b=20$ , $c= 21$ and $sin \ A = \frac{3}{5}$

To Find:

• Value of a

Solution:

Cos A = 4/5      

⇒      $Sin^2 x + Cos^2 x = 1$

⇒     $cos^2 x = 1 - sin^2x$

⇒      $cos^2 x = 1 - ({\frac{3}{5} })^2$

        $cos^2 x = 1 - \frac{9}{25}\\ Cos x = \frac{4}{5}$

In a triangle,

⇒         $a^2 = b^2 + c^2 - 2bc \times cos \ A$

 putting the values, we get

    ⇒          $a^2 = (20)^2 + (21)^2 - 2 \times 20 \times 21 \times cos \ A$

    ⇒          $a^2 = 400 + 441 - 2 20 \times 21 \times \frac{4}{5}$

    ⇒          $a^2 = 841 - 672$

    ⇒          $a^2= 169$

    ⇒          $a = 13$

Thus, the value of $a = 13$