Question

find q.
right triangle with angles 30°, 60°, 90°, side opposite 30° is 17√3 cm, side q is opposite 60°
write your answer in simplest radical form.
blank centimeters

Explanation:

Step1: Identify triangle type and trigonometric ratio

This is a 30-60-90 right triangle. We can use the tangent function for the 30° angle. The tangent of an angle in a right triangle is $\frac{\text{opposite}}{\text{adjacent}}$. For the 30° angle, the opposite side to 30° is the side opposite (let's say the side with length related to q, wait no: wait, the angle of 30°, the side adjacent to 30° is $17\sqrt{3}$ cm, and the side opposite to 30°? Wait no, wait the right angle, so the angles are 30°, 60°, 90°. Let's label the sides: the side opposite 30° is the shortest side, opposite 60° is longer, hypotenuse is longest. Wait, in the triangle, the side given is $17\sqrt{3}$ cm, which is adjacent to the 30° angle (since 30° angle has the adjacent side as $17\sqrt{3}$ and the opposite side as q? Wait no, wait the angle of 30°: the right angle is between the two legs. So angle 30°: one leg is adjacent (let's say the leg with length $17\sqrt{3}$), and the other leg is opposite (q). Wait, no, let's check: in a 30-60-90 triangle, the ratios are: sides are in the ratio $1 : \sqrt{3} : 2$, where the side opposite 30° is the shortest (length $x$), opposite 60° is $x\sqrt{3}$, hypotenuse is $2x$. Wait, alternatively, using trigonometry: $\tan(30^\circ) = \frac{\text{opposite}}{\text{adjacent}}$. Wait, angle 30°: opposite side is the side opposite 30°, which would be the side with length... Wait, the triangle has angles 30°, 60°, 90°. So the side opposite 30° is the leg opposite 30°, let's call that $a$, opposite 60° is $b$, hypotenuse $c$. Then $\tan(30^\circ) = \frac{a}{b}$, $\tan(60^\circ) = \frac{b}{a}$. Wait, but in the diagram, the side given is $17\sqrt{3}$ cm, which is adjacent to the 30° angle (since 30° angle is at the top, so the leg adjacent to 30° is $17\sqrt{3}$ cm, and the leg opposite to 30° is q? Wait no, wait the angle of 30°: the two legs are: one leg is adjacent to 30° (length $17\sqrt{3}$), the other leg is opposite to 30° (length q). Wait, no, actually, let's use the tangent of 30°: $\tan(30^\circ) = \frac{\text{opposite}}{\text{adjacent}} = \frac{q}{17\sqrt{3}}$? Wait no, that can't be. Wait, no, maybe I mixed up. Wait, the angle of 60°: let's use the tangent of 60°. $\tan(60^\circ) = \sqrt{3} = \frac{\text{opposite}}{\text{adjacent}}$. For angle 60°, the opposite side is $17\sqrt{3}$? No, wait, let's look at the diagram again. The triangle has a right angle, angle 30° at the top, angle 60° at the bottom, and the side between 30° and right angle is $17\sqrt{3}$ cm, and the side between 60° and right angle is q? Wait, no, the labels: the side labeled $17\sqrt{3}$ cm is adjacent to the 30° angle (since 30° angle is at the vertex where that side and the hypotenuse meet), and the side labeled q is adjacent to the 60° angle. Wait, maybe better to use the tangent of 30°: $\tan(30^\circ) = \frac{\text{opposite}}{\text{adjacent}}$. The angle 30°: opposite side is q (wait, no, if the right angle is between the two legs, then the leg opposite 30° is the one with length q? Wait, no, let's think again. In a 30-60-90 triangle, the sides are in the ratio: side opposite 30°: $x$, side opposite 60°: $x\sqrt{3}$, hypotenuse: $2x$. So if we consider the angle of 30°, the side adjacent to 30° is the side opposite 60°, which is $x\sqrt{3}$, and the side opposite 30° is $x$. So in this case, the side adjacent to 30° is $17\sqrt{3}$ cm, which should be equal to $x\sqrt{3}$ (since it's opposite 60°). So $x\sqrt{3} = 17\sqrt{3}$, so $x = 17$. Then the side opposite 30° (which is q? Wait no, wait the side opposite 30° is $x$, which would be 17? Wait, no, maybe I got the sides reversed. Wait, no, let's use trigonometry. Let's denote: angle 30°, adjacent side (to 30°) is $17\sqrt{3}$ cm, opposite side (to 30°) is q. Then $\tan(30^\circ) = \frac{q}{17\sqrt{3}}$. But $\tan(30^\circ) = \frac{1}{\sqrt{3}}$. So $\frac{1}{\sqrt{3}} = \frac{q}{17\sqrt{3}}$. Multiply both sides by $17\sqrt{3}$: $q = \frac{17\sqrt{3}}{\sqrt{3}} = 17$? Wait, that can't be. Wait, no, maybe I mixed up adjacent and opposite. Wait, angle 60°: let's use angle 60°. The adjacent side to 60° is q, and the opposite side is $17\sqrt{3}$ cm. Then $\tan(60^\circ) = \frac{\text{opposite}}{\text{adjacent}} = \frac{17\sqrt{3}}{q}$. And $\tan(60^\circ) = \sqrt{3}$. So $\sqrt{3} = \frac{17\sqrt{3}}{q}$. Solving for q: $q = \frac{17\sqrt{3}}{\sqrt{3}} = 17$? Wait, no, that's not right. Wait, no, maybe the side given is the opposite side to 60°. Wait, let's start over. In a right triangle, with angles 30°, 60°, 90°, the sides are: - Opposite 30°: $a$ - Opposite 60°: $a\sqrt{3}$ - Hypotenuse: $2a$ So if the side opposite 60° is $17\sqrt{3}$ cm, then $a\sqrt{3} = 17\sqrt{3}$, so $a = 17$ cm. Then the side opposite 30° is $a = 17$ cm, but wait, in the diagram, q is the side opposite 30°? Wait, no, the diagram: the side labeled $17\sqrt{3}$ is adjacent to 30°? Wait, maybe the triangle is labeled such that the angle of 30° has the adjacent side as $17\sqrt{3}$ and the opposite side as q, but that would mean q is opposite 30°, but in 30-60-90, opposite 30° is the shortest side. Wait, but maybe I made a mistake. Wait, let's use the tangent of 30°: $\tan(30^\circ) = \frac{\text{opposite}}{\text{adjacent}} = \frac{q}{17\sqrt{3}}$. $\tan(30^\circ) = \frac{1}{\sqrt{3}}$, so $\frac{1}{\sqrt{3}} = \frac{q}{17\sqrt{3}}$. Multiply both sides by $17\sqrt{3}$: $q = \frac{17\sqrt{3}}{\sqrt{3}} = 17$. Wait, but that seems too small. Wait, no, maybe the side given is the hypotenuse? Wait, no, the diagram shows a right angle, so the two legs are the ones with right angle between them. Wait, the length given is $17\sqrt{3}$ cm, which is a leg, not the hypotenuse. Wait, maybe the angle of 30°: the side adjacent to 30° is q, and the side opposite is $17\sqrt{3}$? Wait, that would make more sense. Let's try that. So angle 30°: adjacent side is q, opposite side is $17\sqrt{3}$ cm. Then $\tan(30^\circ) = \frac{\text{opposite}}{\text{adjacent}} = \frac{17\sqrt{3}}{q}$. $\tan(30^\circ) = \frac{1}{\sqrt{3}}$, so $\frac{1}{\sqrt{3}} = \frac{17\sqrt{3}}{q}$. Cross-multiplying: $q = 17\sqrt{3} \times \sqrt{3} = 17 \times 3 = 51$. Ah! That makes sense. I had the adjacent and opposite reversed. So angle 30°: the side opposite to 30° is $17\sqrt{3}$ cm, and the side adjacent to 30° is q. So $\tan(30^\circ) = \frac{\text{opposite}}{\text{adjacent}} = \frac{17\sqrt{3}}{q}$. Since $\tan(30^\circ) = \frac{1}{\sqrt{3}}$, we have $\frac{1}{\sqrt{3}} = \frac{17\sqrt{3}}{q}$. Solving for q: multiply both sides by q: $q \times \frac{1}{\sqrt{3}} = 17\sqrt{3}$. Then multiply both sides by $\sqrt{3}$: $q = 17\sqrt{3} \times \sqrt{3} = 17 \times 3 = 51$. Yes, that works. Alternatively, using the 30-60-90 triangle ratios: the side opposite 30° is the shortest side (let's call it $x$), opposite 60° is $x\sqrt{3}$, hypotenuse $2x$. Wait, but in this case, the side opposite 30° would be $x$, and the side opposite 60° would be $x\sqrt{3}$. Wait, but if the side opposite 30° is $17\sqrt{3}$? No, that can't be, because $x\sqrt{3}$ would be longer. Wait, no, I think my initial mistake was in identifying which side is which. Let's re-express: in the triangle, the angle of 30°: the leg opposite to 30° is the side with length $17\sqrt{3}$? No, that would mean the side opposite 60° is longer, but $17\sqrt{3}$ is already a longer length. Wait, no, the correct approach is using trigonometry with the correct adjacent and opposite. So angle 30°: adjacent side is q, opposite side is $17\sqrt{3}$ cm. Then $\tan(30^\circ) = \frac{17\sqrt{3}}{q} = \frac{1}{\sqrt{3}}$. Solving for q: $q = 17\sqrt{3} \times \sqrt{3} = 17 \times 3 = 51$. That makes sense. So the length of q is 51 cm.

Step2: Verify with trigonometric ratio or triangle ratios

Alternatively, using the cotangent of 60°: $\cot(60^\circ) = \frac{\text{adjacent}}{\text{opposite}} = \frac{q}{17\sqrt{3}}$. $\cot(60^\circ) = \frac{1}{\sqrt{3}}$, so $\frac{1}{\sqrt{3}} = \frac{q}{17\sqrt{3}}$. Wait, no, that would give q = 17, which is wrong. Wait, no, cotangent of 60° is $\frac{1}{\sqrt{3}}$, but adjacent to 60° is $17\sqrt{3}$? No, I'm getting confused again. Let's stick to the first correct approach: when angle is 30°, opposite side is $17\sqrt{3}$, adjacent is q. $\tan(30°) = \frac{opposite}{adjacent} = \frac{17\sqrt{3}}{q} = \frac{1}{\sqrt{3}}$. Solving gives q = 51. Yes, that's correct.

Answer:

51