How do you solve the triangle when α = 30.0°, a = 4.53, b = 9.06?

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The law of sines gives us:

`a/sinalpha=b/sinbeta`

`asinbeta=bsinalpha`

`sinbeta=(bsinalpha)/a=((9.06)(sin30^@))/4.53=((9.06)(0.5))/4.53=4.53/4.53=1`

`beta=arcsin(1)=90^@`

`gamma=180^@-(30^@+90^@)=180^@-120^@=60^@`

I will try to solve all of the triangle's sides and angles using the Laws of Sines.

Laws of Sines states that `Sin(alpha)/a=Sin(beta)/b=Sin(gamma)/c`

We are assuming that the variables were labeled like the following:

Let's figure out `beta` using the Laws of Sines.
We have: `sin30/4.53=sinbeta/9.06` We now solve for `beta`
=>`9.06sin30=4.53sinbeta`
=>`9.06sin30=4.53sinbeta`
=>`2sin30=sinbeta`
=>`2(0.5)=sinbeta` We use `arccsin` on both sides.
=>`90°=beta`

We could solve for `gamma` by using the fact that the angles of a triangle always add up to 180°.
Therefore, `gamma=180-30-90`

=>`gamma=60°`

We now see this is the `30-60-90` triangle!

Remember that there is a special relationship between the sides of this right triangle. Look below for more understanding:

Since we know the shortest and the longest side, we see that the missing side is the side opposite to `60`degrees.

We now simply multiply our shortest side by `sqrt3`.
`c=4.53sqrt3`
`c~~7.846`
Our triangle now will look like this: