Question

1. solve for x.

Explanation:

Step1: Identify triangle type and angles

This is a right - triangle with one angle \(30^{\circ}\), so the third angle is \(180 - 90 - 30=60^{\circ}\)? Wait, no, wait. Wait, in a right - triangle, the sum of angles is \(180^{\circ}\). Let's assume that the side opposite \(30^{\circ}\) and adjacent? Wait, no, maybe we are missing the hypotenuse or another side? Wait, maybe there is a misinterpretation. Wait, maybe the triangle is a 30 - 60 - 90 triangle, and we know some side relationships. Wait, maybe the side labeled \(66 + x\) is related to the 30 - 60 - 90 triangle ratios. Wait, 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 say length \(a\)), the side opposite \(60^{\circ}\) is \(a\sqrt{3}\), and the hypotenuse is \(2a\). But wait, maybe the right - angle is between the vertical side and the side \(66 + x\). Wait, maybe the angle of \(30^{\circ}\) is such that the side adjacent to \(30^{\circ}\) is the vertical side, and the side opposite is \(66 + x\)? Wait, no, in a right - triangle, \(\tan(\theta)=\frac{\text{opposite}}{\text{adjacent}}\), \(\sin(\theta)=\frac{\text{opposite}}{\text{hypotenuse}}\), \(\cos(\theta)=\frac{\text{adjacent}}{\text{hypotenuse}}\). Wait, maybe the triangle is a 30 - 60 - 90 triangle, and we assume that the hypotenuse is twice the shorter leg. Wait, maybe there is a mistake, or maybe the other leg is of length, say, let's assume that the side opposite \(30^{\circ}\) is \(66 + x\), and the hypotenuse is \(2(66 + x)\), but we need more info. Wait, no, maybe the triangle is isoceles? No, it's a right - triangle with \(30^{\circ}\). Wait, maybe the sum of angles: in a triangle, sum of angles is \(180^{\circ}\). So right - angle is \(90^{\circ}\), one angle is \(30^{\circ}\), so the third angle is \(180-(90 + 30)=60^{\circ}\). But how does that help with \(x\)? Wait, maybe the problem is that the triangle is a 30 - 60 - 90 triangle, and the side \(66 + x\) is equal to the side opposite \(60^{\circ}\), and the side opposite \(30^{\circ}\) is, say, let's assume that the vertical side is of length \(s\), then the side \(66 + x\) would be \(s\sqrt{3}\), and the hypotenuse \(2s\). But we don't have \(s\). Wait, maybe the problem is that the triangle is a right - triangle, and we are supposed to use the fact that the non - right angles add up to \(90^{\circ}\)? No, \(30^{\circ}+60^{\circ}=90^{\circ}\), which is correct. Wait, maybe there is a typo, or maybe the other side is given as, for example, if the side opposite \(30^{\circ}\) is \(66\), but no, the side is \(66 + x\). Wait, maybe the triangle is a 30 - 60 - 90 triangle, and the side adjacent to \(30^{\circ}\) is \(66 + x\), and the side opposite is, say, let's assume that the hypotenuse is \(132 + 2x\) (since in 30 - 60 - 90, hypotenuse is twice the shorter leg). Wait, no, maybe the problem is that the angle of \(30^{\circ}\) and the right - angle, so the third angle is \(60^{\circ}\), and maybe the side \(66 + x\) is equal to the side that is related to the 30 - 60 - 90 ratio. Wait, maybe I made a mistake. Wait, let's re - examine. The triangle is right - angled, one angle is \(30^{\circ}\), so the angles are \(90^{\circ}\), \(30^{\circ}\), and \(60^{\circ}\). Let's assume that the side labeled \(66 + x\) is the side opposite the \(60^{\circ}\) angle, and the side opposite the \(30^{\circ}\) angle is, say, \(y\), then \(66 + x=y\sqrt{3}\), and hypotenuse \(=2y\). But we don't know \(y\). Wait, maybe the problem is that the triangle is isoceles? No, because one angle is \(30^{\circ}\), right - angle is \(90^{\circ}\), so it's not isoceles. Wait, maybe the problem is that the sum of the angles in the triangle is \(180^{\circ}\), but we already used that. Wait, maybe there is a missing side, like the hypotenuse or another leg. Wait, maybe the original problem had a hypotenuse or another leg, but in the given image, we only see \(30^{\circ}\), right - angle, and \(66 + x\). Wait, maybe the side opposite \(30^{\circ}\) is \(66\), and the side adjacent is \(66 + x\)? No, the label is \(66 + x\) on the side adjacent to \(30^{\circ}\)? Wait, no, the right - angle is at the bottom left, so the sides: vertical side (left), horizontal side (bottom, labeled \(66 + x\)), and hypotenuse (right). The angle at the top left is \(30^{\circ}\), so the angle between the vertical side and the hypotenuse is \(30^{\circ}\). So, in that case, \(\tan(30^{\circ})=\frac{\text{opposite}}{\text{adjacent}}=\frac{66 + x}{\text{vertical side}}\), but we don't know the vertical side. Wait, maybe the vertical side is equal to the hypotenuse's half? No, in 30 - 60 - 90, the side opposite \(30^{\circ}\) is half the hypotenuse. The side opposite \(30^{\circ}\) here would be the horizontal side (\(66 + x\))? Wait, no, the angle at the top left is \(30^{\circ}\), so the side opposite to \(30^{\circ}\) is the horizontal side (\(66 + x\)), and the hypotenuse is twice that, so hypotenuse \(=2(66 + x)\). The vertical side would be \(\sqrt{(2(66 + x))^{2}-(66 + x)^{2}}=(66 + x)\sqrt{3}\). But we still don't have enough info. Wait, maybe the problem is that the triangle is a 30 - 60 - 90 triangle, and the side \(66 + x\) is equal to the side that is \( \frac{1}{\sqrt{3}}\) times the vertical side, but we need more. Wait, maybe there is a mistake in the problem, or maybe I misread it. Wait, maybe the angle is \(60^{\circ}\) instead of \(30^{\circ}\)? No, the image shows \(30^{\circ}\). Wait, maybe the problem is that the sum of the angles in the triangle is \(180^{\circ}\), so \(90+30+(66 + x)=180\)? Wait, no, angles are in degrees, and \(66 + x\) is a length, not an angle. Oh! Wait, that's my mistake. \(66 + x\) is a length, not an angle. So the triangle has angles \(90^{\circ}\), \(30^{\circ}\), and \(60^{\circ}\), and one of the legs is \(66 + x\). Let's assume that the side opposite \(30^{\circ}\) is \(66 + x\), and the hypotenuse is \(2(66 + x)\), and the other leg (adjacent to \(30^{\circ}\)) is \((66 + x)\sqrt{3}\). But we need another side to solve for \(x\). Wait, maybe the hypotenuse is given as, say, \(132\) (since \(2\times66 = 132\)), so if hypotenuse is \(132\), then \(2(66 + x)=132\), so \(66 + x = 66\), so \(x = 0\). But that seems odd. Wait, maybe the side opposite \(60^{\circ}\) is \(66\), and the side opposite \(30^{\circ}\) is \(66 + x\). Then, since \(\tan(60^{\circ})=\frac{\text{opposite to }60^{\circ}}{\text{opposite to }30^{\circ}}=\frac{66}{66 + x}=\sqrt{3}\), then \(66=\sqrt{3}(66 + x)\), \(66 + x=\frac{66}{\sqrt{3}} = 22\sqrt{3}\approx38.1\), which would make \(x\) negative, which doesn't make sense. Wait, maybe the side adjacent to \(30^{\circ}\) is \(66\), and the side opposite is \(66 + x\). Then \(\tan(30^{\circ})=\frac{66 + x}{66}=\frac{1}{\sqrt{3}}\), so \(66 + x=\frac{66}{\sqrt{3}} = 22\sqrt{3}\approx38.1\), \(x = 22\sqrt{3}-66\approx - 33.9\), still negative. Wait, maybe the hypotenuse is \(66\), then the side opposite \(30^{\circ}\) is \(\frac{66}{2}=33\), so \(66 + x = 33\), \(x=-33\), negative. This is confusing. Wait, maybe the problem is that the triangle is a right - triangle, and the two non - right angles are \(30^{\circ}\) and \(60^{\circ}\), and the side \(66 + x\) is equal to the side that is \( \frac{\text{hypotenuse}}{2}\), but we need to assume that the hypotenuse is, say, \(132\), then \(66 + x = 66\), \(x = 0\). Alternatively, maybe the problem has a typo, and the angle is \(60^{\circ}\), then the side opposite \(30^{\circ}\) would be \(66 + x\), and the side opposite \(60^{\circ}\) would be \((66 + x)\sqrt{3}\). But without more info, it's hard. Wait, maybe the original problem was that the triangle is a 30 - 60 - 90 triangle, and the side opposite \(30^{\circ}\) is \(x\), and the side adjacent is \(66\), but in the image, it's \(66 + x\). Wait, maybe I misread the label. Maybe it's \(66 - x\) instead of \(66 + x\)? If it's \(66 - x\), and the side opposite \(30^{\circ}\) is \(66 - x\), and hypotenuse is \(2(66 - x)\), and if the other leg is, say, \(66\) (adjacent to \(30^{\circ}\)), then \(\tan(30^{\circ})=\frac{66 - x}{66}=\frac{1}{\sqrt{3}}\), \(66 - x=\frac{66}{\sqrt{3}} = 22\sqrt{3}\), \(x = 66 - 22\sqrt{3}\approx66 - 38.1 = 27.9\). But the label is \(66 + x\). Alternatively, maybe the angle is \(45^{\circ}\), making it an isoceles right - triangle, so \(66 + x\) would be equal to the other leg, but the angle is \(30^{\circ}\). Wait, maybe the problem is that the sum of the angles is \(180^{\circ}\), but \(66 + x\) is an angle, not a length. Oh! That's a possibility. Maybe \(66 + x\) is the measure of the third angle. So in a triangle, sum of angles is \(180^{\circ}\). So \(90+30+(66 + x)=180\). Let's solve that.

Step1: Set up angle sum equation

The sum of the interior angles of a triangle is \(180^{\circ}\). The triangle has angles \(90^{\circ}\), \(30^{\circ}\), and \(66 + x\) (in degrees). So the equation is:
\[90 + 30+(66 + x)=180\]

Step2: Simplify the left - hand side

First, add the known angles: \(90+30 = 120\), so the equation becomes:
\[120+66 + x=180\]
Then, \(120 + 66=186\), so:
\[186+x = 180\]

Step3: Solve for \(x\)

Subtract \(186\) from both sides of the equation:
\[x=180 - 186\]
\[x=-6\]

Answer:

\(x = - 6\)