Question

set up and finish: 30 - 60 - 90 #1
write an equation that can be used to solve for x. your equation must relate back to the 30 - 60 - 90 special triangle. from the last slide 3, this would be step 2.
solve your equation above for x.

Explanation:

Step1: Recall 30 - 60 - 90 triangle ratio

In a 30 - 60 - 90 triangle, the side lengths are in the ratio \(1:\sqrt{3}:2\). The side opposite the 30° angle is the shortest side, the side opposite the 60° angle is \(\sqrt{3}\) times the shortest side, and the hypotenuse is twice the shortest side. Here, the side of length 19 is opposite the 60° angle, and \(x\) is the hypotenuse. Let the shortest side be \(a\). Then the side opposite 60° is \(a\sqrt{3}\), and the hypotenuse \(x = 2a\). Since \(a\sqrt{3}=19\), then \(a=\frac{19}{\sqrt{3}}\). So the equation relating \(x\) and 19 is \(\sin(30^{\circ})=\frac{19}{x}\times\frac{1}{\sqrt{3}}\times\sqrt{3}\) (using trigonometric ratios \(\sin\theta=\frac{\text{opposite}}{\text{hypotenuse}}\), \(\sin(30^{\circ})=\frac{1}{2}\), \(\sin(60^{\circ})=\frac{\sqrt{3}}{2}\)). A more straightforward equation from the side - length ratio: \(\frac{19}{x}=\sin(60^{\circ})\) or \(x=\frac{19}{\sin(60^{\circ})}\). Also, from the ratio \(x = \frac{2\times19}{\sqrt{3}}\) (since if the side opposite 60° is \(s = 19\), and \(s=\frac{\sqrt{3}}{2}x\)).

Step2: Solve for \(x\)

We know that \(\sin(60^{\circ})=\frac{\sqrt{3}}{2}\). From the equation \(x=\frac{19}{\sin(60^{\circ})}\), substitute \(\sin(60^{\circ})=\frac{\sqrt{3}}{2}\). Then \(x=\frac{19}{\frac{\sqrt{3}}{2}}=\frac{38}{\sqrt{3}}\). Rationalize the denominator: \(x=\frac{38\sqrt{3}}{3}\approx21.93\). Another way, using the side - length ratio: If the side opposite 60° is \(s = 19\), and in 30 - 60 - 90 triangle \(s=\frac{\sqrt{3}}{2}x\). Solving for \(x\), we get \(x=\frac{2\times19}{\sqrt{3}}=\frac{38\sqrt{3}}{3}\).

Answer:

The equation is \(x=\frac{2\times19}{\sqrt{3}}\) (or \(x=\frac{19}{\sin(60^{\circ})}\)), and \(x = \frac{38\sqrt{3}}{3}\approx21.93\) units.