2.2k views
0 votes
I need help solving thisIt’s from my online ACT prep guide 21’-22’

I need help solving thisIt’s from my online ACT prep guide 21’-22’-example-1
User Ohmusama
by
8.1k points

1 Answer

4 votes

Let's start by making a sketch:

To figure out the perimeter, we will need all sides.

Since this is a right triangle, we can use trigonometric ratios to find them.

We know the leg adjacent to the 30° angle, so we can use tangent of 30° to find the other leg and either cosine or sine of 30° to find the hypotenuse afterwards.


\begin{gathered} \tan 30\degree=\frac{\text{opposite leg}}{\text{adjacent leg}}=(BC)/(AC) \\ BC=AC\cdot\tan 30\degree \end{gathered}

Tanget of 30° degree is known to be:


\tan 30\degree=\frac{\sqrt[]{3}}{3}

So:


BC=AC\cdot\tan 30\degree=7\sqrt[]{2}\cdot\frac{\sqrt[]{3}}{3}=\frac{7\sqrt[]{6}}{3}

And the value for cosine of 30° is also known:


\cos 30\degree=\frac{\sqrt[]{3}}{2}

So:


\begin{gathered} \cos 30\degree=\frac{\text{adjacent leg}}{\text{hypotenuse}}=(AC)/(AB) \\ AB=(AC)/(\cos30\degree)=\frac{7\sqrt[]{2}}{\frac{\sqrt[]{3}}{2}}=\frac{14\sqrt[]{2}}{\sqrt[]{3}}=\frac{14\sqrt[]{2}\sqrt[]{3}}{3}=\frac{14\sqrt[]{6}}{3} \end{gathered}

So, the perimeter will be:


\begin{gathered} P=AB+BC+AC \\ P=\frac{14\sqrt[]{6}}{3}+\frac{7\sqrt[]{6}}{3}+7\sqrt[]{2} \\ P=\frac{21\sqrt[]{6}}{3}+7\sqrt[]{2} \\ P=7\sqrt[]{6}+7\sqrt[]{2} \end{gathered}

Which corresponds to the third alternative.

I need help solving thisIt’s from my online ACT prep guide 21’-22’-example-1
User SacredGeometry
by
8.6k points

Related questions

asked May 20, 2023 79.6k views
Rturrado asked May 20, 2023
by Rturrado
7.4k points
1 answer
3 votes
79.6k views
asked Apr 14, 2023 193k views
RET asked Apr 14, 2023
by RET
8.6k points
1 answer
4 votes
193k views
1 answer
1 vote
96.9k views