Question

which three pairs of measurements are possible side lengths for the triangle?
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

Explanation:

Step1: Use trigonometric ratios

In a right - angled triangle \(ABC\) with \(\angle B = 90^{\circ}\), \(\angle A=60^{\circ}\), \(\angle C = 30^{\circ}\). We know the trigonometric ratios: \(\sin A=\frac{BC}{AC}\), \(\cos A=\frac{AB}{AC}\), and the property of a \(30 - 60-90\) triangle: \(BC=\sqrt{3}AB\), \(AC = 2AB\).

Step2: Check each option

• Option 1: \(AB = 4\), \(BC = 4\sqrt{3}\)

If \(AB = 4\), then using \(BC=\sqrt{3}AB\), \(BC=\sqrt{3}\times4 = 4\sqrt{3}\), and \(AC=2AB=8\).

• Option 2: \(BC = 2\sqrt{3}\), \(AC = 2\)

If \(BC = 2\sqrt{3}\), using \(BC=\sqrt{3}AB\), then \(AB=\frac{BC}{\sqrt{3}}=\frac{2\sqrt{3}}{\sqrt{3}} = 2\), and \(AC = 2AB=4
eq2\).

• Option 3: \(AB = 3\), \(AC = 3\sqrt{3}\)

If \(AB = 3\), using \(AC = 2AB\), \(AC=6
eq3\sqrt{3}\).

• Option 4: \(BC = 10\), \(AC = 4\sqrt{3}\)

If \(BC = 10\), using \(BC=\sqrt{3}AB\), \(AB=\frac{10}{\sqrt{3}}=\frac{10\sqrt{3}}{3}\), and \(AC = 2AB=\frac{20\sqrt{3}}{3}
eq4\sqrt{3}\).

• Option 5: \(AB = 7\), \(AC = 14\)

If \(AB = 7\), using \(AC = 2AB\), \(AC=2\times7 = 14\).

• Option 6: \(AB = 11\), \(BC = 11\sqrt{3}\)

If \(AB = 11\), using \(BC=\sqrt{3}AB\), \(BC=\sqrt{3}\times11=11\sqrt{3}\).

Answer:

AB = 4, BC = 4√3; AB = 7, AC = 14; AB = 11, BC = 11√3