Question

find p.
right triangle with angles 30°, 60°, 90°, one leg 9√3 yd, hypotenuse p, right angle marked
write your answer in simplest radical form.
blank box yards
square root button

Explanation:

Step1: Identify triangle type and sides

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 \(30^{\circ}\) is the shortest side (let's call it \(x\)), the side opposite \(60^{\circ}\) is \(x\sqrt{3}\), and the hypotenuse is \(2x\). Here, the side with length \(9\sqrt{3}\) yd is opposite the \(60^{\circ}\) angle (since the right angle is between the other two sides, and the \(30^{\circ}\) angle is at the top, so the side \(9\sqrt{3}\) is opposite \(60^{\circ}\)). Let the side opposite \(30^{\circ}\) be \(a\), the side opposite \(60^{\circ}\) be \(a\sqrt{3}\), and hypotenuse \(p\) (wait, no, wait: Wait, in the triangle, the right angle is at the right, so the angle at the bottom is \(60^{\circ}\), angle at the top is \(30^{\circ}\). So the side adjacent to \(30^{\circ}\) is the side with \(9\sqrt{3}\) yd? Wait, no, let's label the triangle. Let's call the right - angled vertex \(C\), the \(30^{\circ}\) vertex \(A\), and \(60^{\circ}\) vertex \(B\). So \(AC = 9\sqrt{3}\) yd, \(\angle A=30^{\circ}\), \(\angle C = 90^{\circ}\), \(\angle B = 60^{\circ}\). Then, in right - triangle trigonometry, \(\cos(30^{\circ})=\frac{\text{adjacent}}{\text{hypotenuse}}\), where adjacent to \(30^{\circ}\) is \(AC = 9\sqrt{3}\), and hypotenuse is \(p\) (side \(AB\)). Wait, \(\cos(30^{\circ})=\frac{\sqrt{3}}{2}\), and \(\cos(A)=\frac{AC}{AB}\), so \(\cos(30^{\circ})=\frac{9\sqrt{3}}{p}\).

Step2: Solve for \(p\)

We know that \(\cos(30^{\circ})=\frac{\sqrt{3}}{2}\), so \(\frac{\sqrt{3}}{2}=\frac{9\sqrt{3}}{p}\). Cross - multiply: \(p\times\sqrt{3}=2\times9\sqrt{3}\). Divide both sides by \(\sqrt{3}\): \(p = 18\). Wait, alternatively, using the 30 - 60 - 90 triangle ratios. The side opposite \(60^{\circ}\) is \(x\sqrt{3}\), and here the side opposite \(60^{\circ}\) is \(9\sqrt{3}\)? Wait, no, the side opposite \(60^{\circ}\) should be \(x\sqrt{3}\), and the hypotenuse is \(2x\). Wait, maybe I mixed up. Let's re - label: In a 30 - 60 - 90 triangle, the side opposite \(30^{\circ}\) is the shortest side (\(x\)), side opposite \(60^{\circ}\) is \(x\sqrt{3}\), hypotenuse is \(2x\). So if the side opposite \(60^{\circ}\) is \(9\sqrt{3}\), then \(x\sqrt{3}=9\sqrt{3}\), so \(x = 9\). Then the hypotenuse \(p = 2x=18\). That matches the trigonometric approach.

Answer:

\(18\)