Question

#16 find the value of the indicated variable. if your answer is not an integer, express it in simplified radical form. triangle with right angle, 30° angle, hypotenuse? opposite side 10√3, side y adjacent to 30°? y = blank

Explanation:

Step1: Identify triangle type

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

Step2: Determine given side

The side given is \(10\sqrt{3}\), which is opposite the \(60^\circ\) angle (since the right angle is one angle, \(30^\circ\) is another, so the third is \(60^\circ\)). So, if the side opposite \(60^\circ\) is \(x\sqrt{3}\), then \(x\sqrt{3}=10\sqrt{3}\). Solving for \(x\), we divide both sides by \(\sqrt{3}\), getting \(x = 10\).

Step3: Find hypotenuse (y)

The hypotenuse in a 30-60-90 triangle is \(2x\). Since \(x = 10\), the hypotenuse \(y = 2\times10 = 20\)? Wait, no, wait. Wait, the side \(10\sqrt{3}\) is opposite \(60^\circ\), so the side opposite \(30^\circ\) is \(x\), and hypotenuse is \(2x\). Wait, no, let's re - check. In a 30 - 60 - 90 triangle:

• Let the side opposite \(30^\circ\) be \(a\).
• Then the side opposite \(60^\circ\) is \(a\sqrt{3}\).
• The hypotenuse is \(2a\).

In the given triangle, the side with length \(10\sqrt{3}\) is opposite the \(60^\circ\) angle (because the right angle is one, \(30^\circ\) is another, so the angle opposite \(10\sqrt{3}\) is \(60^\circ\)). So we have \(a\sqrt{3}=10\sqrt{3}\), so \(a = 10\). Now, the side \(y\) is the hypotenuse? Wait, no, looking at the triangle, the side labeled \(y\) is adjacent to the \(30^\circ\) angle? Wait, no, the triangle is a right - triangle, with right angle, \(30^\circ\) angle, and \(60^\circ\) angle. The side of length \(10\sqrt{3}\) is the side opposite the \(60^\circ\) angle, the side \(y\) is the hypotenuse? Wait, no, maybe I mixed up. Wait, let's use trigonometry. Let's denote the right angle as \(C\), \(30^\circ\) angle as \(A\), and \(60^\circ\) angle as \(B\). The side opposite \(B\) ( \(60^\circ\)) is \(10\sqrt{3}\), the side opposite \(A\) ( \(30^\circ\)) is, say, \(b\), and the hypotenuse (opposite \(C\)) is \(y\).

Using the sine function: \(\sin(60^\circ)=\frac{\text{opposite}}{\text{hypotenuse}}=\frac{10\sqrt{3}}{y}\). We know that \(\sin(60^\circ)=\frac{\sqrt{3}}{2}\). So \(\frac{\sqrt{3}}{2}=\frac{10\sqrt{3}}{y}\). Cross - multiply: \(y\times\sqrt{3}=2\times10\sqrt{3}\). Divide both sides by \(\sqrt{3}\): \(y = 20\). Wait, but earlier when we thought of the 30 - 60 - 90 ratios, if the side opposite \(60^\circ\) is \(10\sqrt{3}\), then the side opposite \(30^\circ\) is \(10\), and hypotenuse is \(20\). Yes, that's correct.

Wait, maybe my initial confusion was about which side is which. Let's confirm with trigonometry. \(\cos(30^\circ)=\frac{\text{adjacent}}{\text{hypotenuse}}\). The adjacent side to \(30^\circ\) is \(10\sqrt{3}\)? No, wait, no. Wait, in the right - triangle, the sides:

• Hypotenuse: \(y\)
• Angle \(30^\circ\): the side adjacent to \(30^\circ\) is the side opposite \(60^\circ\) (length \(10\sqrt{3}\)), and the side opposite \(30^\circ\) is, let's say, \(a\).

We know that \(\cos(30^\circ)=\frac{\text{adjacent}}{\text{hypotenuse}}=\frac{10\sqrt{3}}{y}\). Since \(\cos(30^\circ)=\frac{\sqrt{3}}{2}\), we have \(\frac{\sqrt{3}}{2}=\frac{10\sqrt{3}}{y}\). Solving for \(y\):

Multiply both sides by \(y\): \(\frac{\sqrt{3}}{2}y=10\sqrt{3}\)

Multiply both sides by \(2\): \(\sqrt{3}y = 20\sqrt{3}\)

Divide both sides by \(\sqrt{3}\): \(y = 20\)

Answer:

\(y = 20\)