Question

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

Find sinA

Answer 1

Since ABC is a right triangle, we have

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

Then

`\sin A=\frac ac=\frac{\sqrt3}3=\frac1{\sqrt3}`

Answer 2

`\text{sin}(A)=(√(3))/(3)`

Explanation:

We have been given that in triangle ABC,
`a=√(3)` and
`b=√(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=(√(3))^2+(√(6))^2`

`c^2=3+6`

`c^2=9`

`c=√(9)=3`

`\text{sin}(A)=(√(3))/(3)`

Therefore,
`\text{sin}(A)=(√(3))/(3)`.