Question

in triangle abc, the measure of angle a is 30° and the measure of angle b is 90°. if the length of side bc is 42 centimeters, what is the length, in centimeters, of side ab?
a 42
b 42√2
c 84
d 42√3

Explanation:

Step1: Identify Triangle Type

Triangle \(ABC\) is a right - triangle with \(\angle B = 90^{\circ}\), \(\angle A=30^{\circ}\), so \(\angle C = 60^{\circ}\). Side \(BC\) is opposite \(\angle A\) (opposite \(30^{\circ}\) angle), side \(AB\) is adjacent to \(\angle A\) and opposite \(\angle C\) (opposite \(60^{\circ}\) angle). In a \(30 - 60 - 90\) triangle, the sides are in the ratio \(1:\sqrt{3}:2\) (opposite \(30^{\circ}\), opposite \(60^{\circ}\), hypotenuse respectively). Let \(BC=a = 42\) (opposite \(30^{\circ}\)), \(AB = b\) (opposite \(60^{\circ}\)). The ratio of sides opposite \(30^{\circ}\) to opposite \(60^{\circ}\) is \(\frac{a}{b}=\frac{1}{\sqrt{3}}\).

Step2: Solve for \(AB\)

We know \(a = 42\), from \(\frac{a}{b}=\frac{1}{\sqrt{3}}\), we can re - arrange to get \(b=a\sqrt{3}\). Substitute \(a = 42\) into the formula, we have \(b = 42\sqrt{3}\)? Wait, no, wait. Wait, in right - triangle \(ABC\), \(\tan A=\frac{BC}{AB}\). Since \(\angle A = 30^{\circ}\), \(\tan30^{\circ}=\frac{BC}{AB}\), \(\tan30^{\circ}=\frac{1}{\sqrt{3}}\), \(BC = 42\). So \(\frac{1}{\sqrt{3}}=\frac{42}{AB}\), then \(AB = 42\sqrt{3}\)? Wait, no, \(\tan A=\frac{\text{opposite}}{\text{adjacent}}=\frac{BC}{AB}\). \(\angle A = 30^{\circ}\), opposite side to \(A\) is \(BC\), adjacent side is \(AB\). So \(\tan(30^{\circ})=\frac{BC}{AB}\), \(\tan(30^{\circ})=\frac{1}{\sqrt{3}}\), so \(AB=\frac{BC}{\tan(30^{\circ})}=\frac{42}{\frac{1}{\sqrt{3}}}=42\sqrt{3}\). Alternatively, using the \(30 - 60 - 90\) triangle ratios: In a \(30 - 60 - 90\) triangle, 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 \(2x\). Here, \(BC\) is opposite \(30^{\circ}\) (since \(\angle A = 30^{\circ}\) and \(\angle B = 90^{\circ}\), so side opposite \(\angle A\) is \(BC\)), so \(BC=x = 42\). The side opposite \(60^{\circ}\) (which is \(\angle C\)) is \(AB\), so \(AB=x\sqrt{3}=42\sqrt{3}\). Wait, but let's check with trigonometry again. \(\cos A=\frac{AB}{AC}\), \(\sin A=\frac{BC}{AC}\). But maybe using tangent is better. \(\tan A=\frac{BC}{AB}\), so \(AB=\frac{BC}{\tan A}\). \(\tan30^{\circ}=\frac{1}{\sqrt{3}}\), so \(AB = 42\div\frac{1}{\sqrt{3}}=42\sqrt{3}\).

Wait, but let's re - examine the triangle. \(\angle B = 90^{\circ}\), so \(AC\) is the hypotenuse. \(\angle A = 30^{\circ}\), so \(BC=\frac{1}{2}AC\) (since in a right - triangle, the side opposite \(30^{\circ}\) is half the hypotenuse). So \(AC = 84\). Then using Pythagoras theorem: \(AB=\sqrt{AC^{2}-BC^{2}}=\sqrt{84^{2}-42^{2}}=\sqrt{(84 - 42)(84 + 42)}=\sqrt{42\times126}=\sqrt{42\times42\times3}=42\sqrt{3}\). Yes, that's correct.

Answer:

D. \(42\sqrt{3}\)