Question

question 9 of 10
in the 30 - 60 - 90 triangle below, side s has a length of __ and the hypotenuse has a length of __.
a. \\( \sqrt { 3 } ; 2 \\)
b. \\( 1 ; \sqrt { 2 } \\)
c. \\( 1 ; 1.7 \\)
d. \\( \sqrt { 2 } ; 2 \\)
e. \\( 1, \sqrt { 3 } \\)
f. \\( 2 ; 1.4 \\)

Explanation:

Step1: Recall 30-60-90 triangle ratios

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 (\(x\)), opposite \(60^\circ\) is \(x\sqrt{3}\), and hypotenuse is \(2x\). Here, the side opposite \(60^\circ\) is 1? Wait, no—wait, the right angle is at the bottom right, \(60^\circ\) at bottom left, \(30^\circ\) at top. So the side opposite \(30^\circ\) is the shortest leg (let's say length \(x\)), opposite \(60^\circ\) is \(x\sqrt{3}\), hypotenuse \(2x\). Wait, the side labeled "1" is opposite \(30^\circ\)? Wait no, angle at bottom left is \(60^\circ\), right angle at bottom right, so the side adjacent to \(60^\circ\) (the leg) is... Wait, let's label the triangle: vertices: let’s say \(A\) (top, \(30^\circ\)), \(B\) (bottom right, \(90^\circ\)), \(C\) (bottom left, \(60^\circ\)). So side \(BC = 1\) (adjacent to \(60^\circ\), opposite \(30^\circ\)? Wait, angle at \(A\) is \(30^\circ\), so side opposite \(30^\circ\) is \(BC = 1\). Then the side opposite \(60^\circ\) (which is \(AC\), labeled \(s\)) should be \(1 \times \sqrt{3}=\sqrt{3}\)? Wait no, wait: in 30-60-90, sides: opposite 30°: \(x\), opposite 60°: \(x\sqrt{3}\), hypotenuse: \(2x\). If the side opposite 30° (angle at \(A\)) is \(BC = 1\) (since angle at \(A\) is 30°, so side opposite is \(BC\)), then:

- Side opposite 30° (\(BC\)): \(x = 1\)
- Side opposite 60° (\(AC = s\)): \(x\sqrt{3}=1 \times \sqrt{3}=\sqrt{3}\)? Wait no, wait angle at \(C\) is 60°, so side opposite 60° is \(AB\)? Wait, I'm confused. Wait, angle at \(B\) is 90°, so sides: \(AB\) (hypotenuse, \(h\)), \(BC\) (leg, length 1), \(AC\) (leg, length \(s\)). Angle at \(C\) is 60°, so in triangle \(ABC\), angle at \(C\) is 60°, angle at \(A\) is 30°, angle at \(B\) is 90°. So:

- \(\tan(60^\circ)=\frac{AB}{BC}\)? No, \(\tan(\theta)=\frac{\text{opposite}}{\text{adjacent}}\). Angle at \(C\) is 60°, so opposite side is \(AB\)? No, angle at \(C\) is 60°, so opposite side is \(AB\) (length \(h\)), adjacent is \(BC = 1\). Wait, \(\tan(60^\circ)=\frac{AB}{BC}\)? No, \(\tan(60^\circ)=\frac{\text{opposite}}{\text{adjacent}}=\frac{AB}{BC}\)? Wait, \(AB\) is opposite angle \(C\) (60°), \(BC\) is adjacent to angle \(C\). So \(\tan(60^\circ)=\frac{AB}{BC}\)? No, \(AB\) is the hypotenuse? Wait no, \(AB\) is a leg? Wait, no: \(B\) is the right angle, so legs are \(BC\) and \(AC\), hypotenuse is \(AB\) (labeled \(h\)). So:

- Leg \(BC = 1\) (adjacent to angle \(C\) (60°), opposite angle \(A\) (30°))
- Leg \(AC = s\) (adjacent to angle \(A\) (30°), opposite angle \(C\) (60°))
- Hypotenuse \(AB = h\)

So using trigonometric ratios:

For angle \(C\) (60°):

\(\tan(60^\circ)=\frac{\text{opposite}}{\text{adjacent}}=\frac{AC}{BC}\) → \(\tan(60^\circ)=\frac{s}{1}\) → \(s = \tan(60^\circ)=\sqrt{3}\)

\(\sin(60^\circ)=\frac{\text{opposite}}{\text{hypotenuse}}=\frac{AC}{h}\) → \(\sin(60^\circ)=\frac{\sqrt{3}}{h}\) → \(h=\frac{\sqrt{3}}{\sin(60^\circ)}\). But \(\sin(60^\circ)=\frac{\sqrt{3}}{2}\), so \(h=\frac{\sqrt{3}}{\frac{\sqrt{3}}{2}} = 2\)

Alternatively, using 30-60-90 ratios: the side opposite 30° (which is \(BC = 1\)) is the shortest leg (\(x = 1\)). Then:

- Side opposite 60° (\(AC = s\)): \(x\sqrt{3}=1 \times \sqrt{3}=\sqrt{3}\)? Wait no, wait earlier calculation with tan(60°) gave \(s = \sqrt{3}\), and hypotenuse \(h = 2x = 2 \times 1 = 2\). So side \(s\) (which is \(AC\)) is \(\sqrt{3}\), hypotenuse \(h\) is 2. So the first blank (side \(s\)) is \(\sqrt{3}\), second (hypotenuse) is 2. Looking at options, option A is \(\sqrt{3}; 2\).

Step2: Verify with options

Option A: \(\sqrt{3}\) (side \(s\)) and 2 (hypotenuse). Which matches our calculation.

Answer:

A. \(\sqrt{3}\); 2