# How do you find the remaining side of a 30^circ-60^circ-90^circ triangle if the side opposite 60^circ is 6?

Feb 24, 2017

Use Trigonometric identities.

#### Explanation:

Let us assume the side next to 60° is $a$, and the hypotenuse is $b$.

Then use the Pythagorean theorem.

$b = \sqrt{{6}^{2} + {a}^{2}}$

We know: " "sin 60° = sqrt3/2
Then:
$\frac{6}{\sqrt{{6}^{2} + {a}^{2}}} = \frac{\sqrt{3}}{2}$
${6}^{2} + {a}^{2} = {\left(\frac{6 \times 2}{\sqrt{3}}\right)}^{2} = 48$
${a}^{2} = 12$
$a = 2 \sqrt{3}$
$b = 4 \sqrt{3}$

Feb 24, 2017

The side lengths are: $2 \sqrt{3} , \text{ "6," } 4 \sqrt{3}$.

#### Explanation:

The sides of a 30°"-"60°"-"90° triangle are always of the ratio $1 \text{-"sqrt3"-} 2$. Meaning: the side opposite 60° is $\sqrt{3}$ times the length of the side opposite 30°, and the side opposite 90° is $2$ times as long as the side opposite 30°.

In math:

$\text{side opposite 60°"/"side opposite 30°} = \frac{\sqrt{3}}{1} = \sqrt{3}$

$\text{side opposite 90°"/"side opposite 30°} = \frac{2}{1} = 2$

We are given the side opposite 60° to be length 6. So, given that the ratio of "the 60° side"-to-"the 30° side" is $\sqrt{3} \text{-to-1}$, we can solve:

$\text{side opp. 60°"/"side opp. 30°} = \sqrt{3}$

$\frac{6}{\text{side opp. 30°}} = \sqrt{3}$

$\text{ "6/sqrt3" "="side opp. 30°}$

$\text{ "(6sqrt3)/3" "="side opp. 30°}$

$\text{ "2sqrt3" "="side opp. 30°}$

And, since "the 90° side" is 2 times as long as "the 30° side", we have

$\text{side opp. 90°" = 2xx "side opp. 30°}$
$\text{side opp. 90°} = 2 \times 2 \sqrt{3}$
$\text{side opp. 90°} = 4 \sqrt{3}$.