Question

find t.
write your answer in simplest radical form.

units

Explanation:

Step1: Identify the triangle type

This is a 30-60-90 right triangle. In a 30-60-90 triangle, the sides are in the ratio \(1 : \sqrt{3} : 2\), where the side opposite 30° is the shortest one (let's call it \(x\)), the side opposite 60° is \(x\sqrt{3}\), and the hypotenuse is \(2x\).

Step2: Determine which side is given

The side given is \(17\sqrt{3}\), which is opposite the 60° angle (since the right angle is 90°, the 30° angle's opposite side is \(t\)? Wait, no: Wait, the right angle is at the top, so the side adjacent to 30° is \(t\)? Wait, no, let's label the triangle. Let's call the right angle \(C\), the 30° angle \(A\), and the 60° angle \(B\). Then, side opposite \(A\) (30°) is \(BC = t\), side opposite \(B\) (60°) is \(AC = 17\sqrt{3}\), and hypotenuse \(AB\).

In a 30-60-90 triangle, \(\tan(30^\circ)=\frac{BC}{AC}\), but maybe easier to use ratios. Wait, the side opposite 60° is \(x\sqrt{3}\), and the side opposite 30° is \(x\). Wait, here \(AC = 17\sqrt{3}\) is opposite 60°, so \(x\sqrt{3}=17\sqrt{3}\), so \(x = 17\). Then the side opposite 30° (which is \(BC = t\)) is \(x = 17\)? Wait, no, wait: Wait, in 30-60-90, the sides: hypotenuse is \(2x\), side opposite 30° is \(x\), side opposite 60° is \(x\sqrt{3}\). So if the side opposite 60° is \(17\sqrt{3}\), then \(x\sqrt{3}=17\sqrt{3}\), so \(x = 17\). Then the side opposite 30° (which is \(t\)) is \(x = 17\)? Wait, no, wait the right angle is at the top, so the vertical side is \(t\), horizontal side is \(17\sqrt{3}\), angle at the left is 30°. So angle 30°: adjacent side is \(17\sqrt{3}\), opposite side is \(t\). Wait, maybe using tangent: \(\tan(30^\circ)=\frac{t}{17\sqrt{3}}\). But \(\tan(30^\circ)=\frac{1}{\sqrt{3}}\), so \(\frac{1}{\sqrt{3}}=\frac{t}{17\sqrt{3}}\), multiply both sides by \(17\sqrt{3}\): \(t = \frac{17\sqrt{3}}{\sqrt{3}} = 17\). Wait, that's simpler. Alternatively, since in 30-60-90, the side opposite 30° is half the hypotenuse, but maybe the other way: the side adjacent to 30° is \(x\sqrt{3}\), and opposite is \(x\). Wait, maybe I mixed up. Let's re-express:

In a 30-60-90 triangle, the sides are:

• Opposite 30°: \(x\)
• Opposite 60°: \(x\sqrt{3}\)
• Hypotenuse: \(2x\)

Here, the side with length \(17\sqrt{3}\) is opposite the 60° angle (since the angle at the bottom is 60°), so that side is \(x\sqrt{3}\). So \(x\sqrt{3}=17\sqrt{3}\), so \(x = 17\). The side opposite the 30° angle (which is \(t\)) is \(x\), so \(t = 17\). Wait, but let's check with tangent. \(\tan(30^\circ)=\frac{\text{opposite}}{\text{adjacent}}=\frac{t}{17\sqrt{3}}\). \(\tan(30^\circ)=\frac{1}{\sqrt{3}}\), so \(\frac{1}{\sqrt{3}}=\frac{t}{17\sqrt{3}}\). Multiply both sides by \(17\sqrt{3}\): \(t = \frac{17\sqrt{3}}{\sqrt{3}} = 17\). Yep, that works.

Answer:

\(17\)