select all the correct answers. which three pairs of measurements are possible side lengths for the triangle? triangle image with right angle at b, angle at a is 60°, angle at c is 30° options: ab = 4, bc = 4√3; bc = 2√3, ac = 2; ab = 3, ac = 3√3; bc = 10, ac = 4√3; ab = 7, ac = 14; ab = 11, bc = 11√3

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Accepted Answer

# Explanation:
## Step1: Recall 30-60-90 triangle ratios
In a 30-60-90 right triangle, the sides are in the ratio $1 : \sqrt{3} : 2$, where:
- The side opposite $30^\circ$ (shorter leg, $AB$ here as $\angle C = 30^\circ$) is $x$,
- The side opposite $60^\circ$ (longer leg, $BC$ here as $\angle A = 60^\circ$) is $x\sqrt{3}$,
- The hypotenuse ($AC$ here) is $2x$.

## Step2: Analyze each option
- **Option 1: $AB = 4$, $BC = 4\sqrt{3}$, $AC = 8$? Wait, no, $AC$ should be $2x$. Wait, $AB = x = 4$, so $BC = 4\sqrt{3}$, $AC = 8$? Wait, no, the option says $AC = 4\sqrt{3}$? Wait, no, original option: $AB = 4$, $BC = 4\sqrt{3}$ – Wait, no, let's recheck. Wait, the triangle has $\angle A = 60^\circ$, $\angle C = 30^\circ$, right angle at $B$. So:
  - $AB$ is adjacent to $60^\circ$, opposite to $30^\circ$ (wait, no: $\angle A = 60^\circ$, so $AB$ is adjacent to $\angle A$, $BC$ is opposite to $\angle A$. Wait, maybe better to use trigonometry:
    - $\sin 60^\circ = \frac{BC}{AC}$, $\cos 60^\circ = \frac{AB}{AC}$, $\tan 60^\circ = \frac{BC}{AB}$
    - $\cos 60^\circ = 0.5 = \frac{AB}{AC} \implies AC = 2AB$
    - $\tan 60^\circ = \sqrt{3} = \frac{BC}{AB} \implies BC = AB\sqrt{3}$
    - $\sin 60^\circ = \frac{\sqrt{3}}{2} = \frac{BC}{AC} \implies BC = \frac{\sqrt{3}}{2}AC$

Let's check each option:
  1. $AB = 4$, $BC = 4\sqrt{3}$, $AC = 8$? Wait, the option says $AC = 4\sqrt{3}$? No, the first option is $AB = 4$, $BC = 4\sqrt{3}$ – Wait, no, the user's option 1: $AB = 4$, $BC = 4\sqrt{3}$ – Wait, let's use the ratio. If $AB = x$, then $BC = x\sqrt{3}$, $AC = 2x$.
    - For $AB = 4$ (so $x = 4$): $BC = 4\sqrt{3}$, $AC = 8$. But the option's $AC$ is not given? Wait, no, the first option is $AB = 4$, $BC = 4\sqrt{3}$ – Wait, maybe the option was miswritten? Wait, the first option: $AB = 4$, $BC = 4\sqrt{3}$ – Let's check $\tan 60^\circ = \frac{BC}{AB} = \frac{4\sqrt{3}}{4} = \sqrt{3}$, which is correct. And $AC$ should be $2AB = 8$, but the option doesn't list $AC$? Wait, no, the first option is $AB = 4$, $BC = 4\sqrt{3}$ – Wait, maybe the option is $AB = 4$, $BC = 4\sqrt{3}$, $AC = 8$? But the given option is $AB = 4$, $BC = 4\sqrt{3}$ – Wait, maybe the user's first option is $AB = 4$, $BC = 4\sqrt{3}$ (and $AC = 8$, but the option doesn't show $AC$? No, the original options:

Wait, the options are:
  1. $AB = 4$, $BC = 4\sqrt{3}$ (wait, no, the first option is $AB = 4$, $BC = 4\sqrt{3}$ – Let's check the ratio: $AB = x$, $BC = x\sqrt{3}$, so $x = 4$, $BC = 4\sqrt{3}$, $AC = 8$. But the option as written: $AB = 4$, $BC = 4\sqrt{3}$ – Maybe the $AC$ is missing? No, the first option is $AB = 4$, $BC = 4\sqrt{3}$ – Wait, maybe I misread. Let's check the second option: $BC = 2\sqrt{3}$, $AC = 2$, $AB =?$

- **Option 2: $BC = 2\sqrt{3}$, $AC = 2$, $AB =?$** Using $\sin 60^\circ = \frac{BC}{AC} \implies \frac{\sqrt{3}}{2} = \frac{2\sqrt{3}}{2} = \sqrt{3}$, which is false. So this is wrong.

- **Option 3: $AB = 3$, $AC = 3\sqrt{3}$** Using $\cos 60^\circ = \frac{AB}{AC} \implies 0.5 = \frac{3}{3\sqrt{3}} = \frac{1}{\sqrt{3}} \approx 0.577$, not 0.5. Wrong.

- **Option 4: $BC = 10$, $AC = 4\sqrt{3}$** $\sin 60^\circ = \frac{10}{4\sqrt{3}} \approx \frac{10}{6.928} \approx 1.443$, which is more than 1. Impossible. Wrong.

- **Option 5: $AB = 7$, $AC = 14$** $\cos 60^\circ = \frac{7}{14} = 0.5$, correct. Then $BC = AB\sqrt{3} = 7\sqrt{3}$? Wait, no, $BC = \sqrt{AC^2 - AB^2} = \sqrt{14^2 - 7^2} = \sqrt{196 - 49} = \sqrt{147} = 7\sqrt{3}$. But the option doesn't list $BC$, but the question is about pairs. Wait, the option is $AB = 7$, $AC = 14$ – since $AC = 2AB$, this fits the 30-60-90 ratio (as $AC$ is hypotenuse, $AB$ is shorter leg? Wait, no: $\angle C = 30^\circ$, so the side opposite $30^\circ$ is $AB$ (since $\angle C = 30^\circ$, opposite side is $AB$), so $AB = \frac{1}{2}AC$, which is true here ($7 = \frac{1}{2} \times 14$). Then $BC = AB\sqrt{3} = 7\sqrt{3}$, but the option is just $AB = 7$, $AC = 14$ – this is a valid pair.

- **Option 6: $AB = 11$, $BC = 11\sqrt{3}$** $\tan 60^\circ = \frac{BC}{AB} = \frac{11\sqrt{3}}{11} = \sqrt{3}$, correct. And $AC = 2AB = 22$ (not listed, but the pair $AB = 11$, $BC = 11\sqrt{3}$ fits the ratio $BC = AB\sqrt{3}$).

Wait, let's re-express the options correctly:

- Option 1: $AB = 4$, $BC = 4\sqrt{3}$ – $BC = AB\sqrt{3}$ (since $4\sqrt{3} = 4 \times \sqrt{3}$), so this is correct (hypotenuse $AC$ would be $8$, but the option doesn't list $AC$, but the pair $AB, BC$ is correct).

- Option 2: $BC = 2\sqrt{3}$, $AC = 2$ – $\sin 60^\circ = \frac{BC}{AC} = \frac{2\sqrt{3}}{2} = \sqrt{3}$, but $\sin 60^\circ = \frac{\sqrt{3}}{2}$, so wrong.

- Option 3: $AB = 3$, $AC = 3\sqrt{3}$ – $\cos 60^\circ = \frac{AB}{AC} = \frac{3}{3\sqrt{3}} = \frac{1}{\sqrt{3}}
eq 0.5$, wrong.

- Option 4: $BC = 10$, $AC = 4\sqrt{3}$ – $\sin 60^\circ = \frac{10}{4\sqrt{3}} \approx 1.44 > 1$, impossible, wrong.

- Option 5: $AB = 7$, $AC = 14$ – $AC = 2AB$ (since $14 = 2 \times 7$), which fits the 30-60-90 ratio (hypotenuse is twice the shorter leg, $AB$ is shorter leg as it's opposite $30^\circ$), correct.

- Option 6: $AB = 11$, $BC = 11\sqrt{3}$ – $BC = AB\sqrt{3}$ (since $11\sqrt{3} = 11 \times \sqrt{3}$), correct (hypotenuse $AC = 22$, not listed, but the pair $AB, BC$ is correct).

Wait, the original options (from the image):

1. $AB = 4$, $BC = 4\sqrt{3}$ – correct (ratio $BC = AB\sqrt{3}$)
  2. $BC = 2\sqrt{3}$, $AC = 2$ – wrong (as above)
  3. $AB = 3$, $AC = 3\sqrt{3}$ – wrong (should be $AC = 2AB = 6$, or $BC = 3\sqrt{3}$)
  4. $BC = 10$, $AC = 4\sqrt{3}$ – wrong (impossible)
  5. $AB = 7$, $AC = 14$ – correct ( $AC = 2AB$)
  6. $AB = 11$, $BC = 11\sqrt{3}$ – correct ( $BC = AB\sqrt{3}$)

Wait, but the first option: $AB = 4$, $BC = 4\sqrt{3}$ – let's check hypotenuse $AC$: $AC = \sqrt{AB^2 + BC^2} = \sqrt{16 + 48} = \sqrt{64} = 8$, which is $2 \times 4$, so that's correct (30-60-90). So:

- Correct options: 1 ( $AB = 4$, $BC = 4\sqrt{3}$ ), 5 ( $AB = 7$, $AC = 14$ ), 6 ( $AB = 11$, $BC = 11\sqrt{3}$ )

Wait, but let's recheck option 1: the option says $AB = 4$, $BC = 4\sqrt{3}$ – yes, that's correct. Option 5: $AB = 7$, $AC = 14$ – correct (since $AC = 2AB$). Option 6: $AB = 11$, $BC = 11\sqrt{3}$ – correct (since $BC = AB\sqrt{3}$).

Wait, but the original options as per the image:

- First option: $AB = 4$, $BC = 4\sqrt{3}$ (correct)
  - Second: $BC = 2\sqrt{3}$, $AC = 2$ (wrong)
  - Third: $AB = 3$, $AC = 3\sqrt{3}$ (wrong)
  - Fourth: $BC = 10$, $AC = 4\sqrt{3}$ (wrong)
  - Fifth: $AB = 7$, $AC = 14$ (correct)
  - Sixth: $AB = 11$, $BC = 11\sqrt{3}$ (correct)

So the correct options are:

- $AB = 4$, $BC = 4\sqrt{3}$ (first option)
  - $AB = 7$, $AC = 14$ (fifth option)
  - $AB = 11$, $BC = 11\sqrt{3}$ (sixth option)

# Answer:
- The correct options are:
  - $AB = 4$, $BC = 4\sqrt{3}$
  - $AB = 7$, $AC = 14$
  - $AB = 11$, $BC = 11\sqrt{3}$

(Note: Based on the 30-60-90 triangle side ratios $x : x\sqrt{3} : 2x$, these pairs satisfy the trigonometric relationships for the given angles.)

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QUESTION IMAGE

Question

Explanation:

Step1: Recall 30-60-90 triangle ratios

Step2: Analyze each option

Answer: