Question

find w.
image of a right triangle with angles 30°, 60°, 90°, one leg 7 km, hypotenuse w
write your answer in simplest radical form.
blank kilometers

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 $30^{\circ}$ is the shortest side (let's call it $x$), the side opposite $60^{\circ}$ is $x\sqrt{3}$, and the hypotenuse is $2x$.

Step2: Determine the side opposite $30^{\circ}$

The side of length $7$ km is opposite the $30^{\circ}$ angle? Wait, no. Wait, the right angle is between the side of length $7$ km and the hypotenuse? Wait, no. Let's label the triangle. Let the right angle be $C$, the $30^{\circ}$ angle be $A$, and the $60^{\circ}$ angle be $B$. Then side opposite $A$ (30°) is $BC = 7$ km, side opposite $B$ (60°) is $AC$, and hypotenuse $AB = w$.
In a 30 - 60 - 90 triangle, $\sin(30^{\circ})=\frac{\text{opposite}}{\text{hypotenuse}}=\frac{7}{w}$, but wait, $\sin(30^{\circ})=\frac{1}{2}$, so $\frac{1}{2}=\frac{7}{w}$ would give $w = 14$, but that's not right because the side opposite 60° should be longer. Wait, I made a mistake. Let's use cosine. The side adjacent to the 60° angle is 7 km, and the hypotenuse is $w$. So $\cos(60^{\circ})=\frac{\text{adjacent}}{\text{hypotenuse}}=\frac{7}{w}$. But $\cos(60^{\circ})=\frac{1}{2}$, so $\frac{1}{2}=\frac{7}{w}\implies w = 14$? No, that's not matching. Wait, no, the side opposite 30° is 7 km? Wait, the angles: 30°, 60°, 90°. So the side opposite 30° is the shortest side. Let's check the angles. The angle at the top is 30°, the right angle is at the right, and the bottom angle is 60°. So the side opposite 30° (top angle) is the vertical side, which is 7 km? Wait, no, the vertical side is adjacent to the 60° angle. Wait, let's use sine of 60°. $\sin(60^{\circ})=\frac{\text{opposite}}{\text{hypotenuse}}$. The opposite side to 60° is the horizontal side, and the hypotenuse is $w$. The adjacent side to 60° is 7 km. Wait, maybe using tangent. $\tan(60^{\circ})=\frac{\text{opposite}}{\text{adjacent}}=\frac{\text{opposite}}{7}$. Since $\tan(60^{\circ})=\sqrt{3}$, then opposite side $=\ 7\sqrt{3}$. Then using Pythagoras: $w^{2}=7^{2}+(7\sqrt{3})^{2}=49 + 147 = 196$. No, that gives $w = 14$, which is wrong. Wait, I think I mixed up the angles. Let's start over. In a 30 - 60 - 90 triangle, the sides are in the ratio $x:x\sqrt{3}:2x$, where $x$ is the side opposite 30°, $x\sqrt{3}$ opposite 60°, and $2x$ hypotenuse. So if the side opposite 30° is $x$, then hypotenuse is $2x$, and side opposite 60° is $x\sqrt{3}$. Now, looking at the triangle, the side of length 7 km: which angle is it opposite to? The angle of 30°: if the side opposite 30° is 7 km, then hypotenuse $w = 2x=14$, and side opposite 60° is $7\sqrt{3}$. But that doesn't match the labels. Wait, maybe the side of 7 km is opposite 60°? No, 60° opposite side should be longer than 30° opposite. Wait, I think the correct approach is: the triangle has angles 30°, 60°, 90°, so it's a 30 - 60 - 90 triangle. The side adjacent to 30° is 7 km? No, let's use cosine of 30°. $\cos(30^{\circ})=\frac{\text{adjacent}}{\text{hypotenuse}}=\frac{7}{w}$. Since $\cos(30^{\circ})=\frac{\sqrt{3}}{2}$, then $\frac{\sqrt{3}}{2}=\frac{7}{w}\implies w=\frac{14}{\sqrt{3}}=\frac{14\sqrt{3}}{3}$. Ah, that's it. So the adjacent side to 30° is 7 km, hypotenuse is $w$. So $\cos(30^{\circ})=\frac{7}{w}$, so $w=\frac{7}{\cos(30^{\circ})}=\frac{7}{\frac{\sqrt{3}}{2}}=\frac{14\sqrt{3}}{3}$.

Answer:

$\frac{14\sqrt{3}}{3}$