Question

In right triangle ABC, a= sqrt3 and b= sqrt 6.Find sinA

Answer 1

Answer:

$\text{sin}(A)=\frac{\sqrt{3}}{3}$

Step-by-step explanation:

We have been given that in triangle ABC, $a=\sqrt{3}$ and $b=\sqrt{6}$. We are asked to find sin of angle A.

We know that sine relates opposite side of right triangle to hypotenuse.

$\text{sin}=\frac{\text{Opposite}}{\text{Hypotenuse}}$

We will use Pythagoras theorem to find the length of hypotenuse.

$c^2=a^2+b^2$

$c^2=(\sqrt{3})^2+(\sqrt{6})^2$

$c^2=3+6$

$c^2=9$

$c=\sqrt{9}=3$

$\text{sin}(A)=\frac{\sqrt{3}}{3}$

Therefore, $\text{sin}(A)=\frac{\sqrt{3}}{3}$.

Answer 2

Since ABC is a right triangle, we have

$a^2+b^2=3+6=9=c^2\implies c=3$

Then

$\sin A=\dfrac ac=\dfrac{\sqrt3}3=\dfrac1{\sqrt3}$