Question

Sketch a 30-60-90 triangle. If the longest side of a similar triangle one foot, then how many inches is the shortest side?

Answer 1

Given that the angles of the triangle are 30, 60, and 90.

Using the sine law

$\frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}$

Substitute A=30, B=60, and C=90, we get

$\frac{a}{\sin 30^o}=\frac{b}{\sin 60^o}=\frac{c}{\sin 90^o}$$\text{Substitute }\sin 30^o=\frac{1}{2},\sin 60^o=\frac{\sqrt[]{3}}{2},\text{ and }\sin 90^o=1,\text{ we get}$

$\frac{a}{\frac{1}{2}}=\frac{b}{\frac{\sqrt[]{3}}{2}}=\frac{c}{1}$

$2a=\frac{2b}{\sqrt[]{3}}=c$

We know that the longest side should be the opposite side of the big angle (90).

Let c be the longest side of the given triangle.

Substitute c=1, we get

$2a=\frac{2b}{\sqrt[]{3}}=1$$2a=1\text{ and }\frac{2b}{\sqrt[]{3}}=1$

$a=\frac{1}{2}\text{ and }b=\frac{\sqrt[]{3}}{2}$$a=0.5\text{ and b=}0.866$

The smallest side is 0.5 feet

Conver the feet into inches.

$1\text{ foo}t\text{ =12 inches }$

Dividing both sides by 2, we get

$\frac{1}{2}\text{ foo}t\text{ =}\frac{12}{2}\text{ inches }$

$0.5\text{foo}t\text{ =}6\text{ inches }$

Hence the smallest side is 6 inches.