Question

6 practice 6 (from unit 4, lesson 3) what is the length of side ab? do not round your answer. type your answer in the box. ______ units how did i do?

Explanation:

Step1: Identify triangle type and trigonometric ratio

We have a right - triangle \(ABC\) with \(\angle A = 90^{\circ}\), \(\angle C=30^{\circ}\) and \(AC = 6\). We want to find the length of \(AB\). We can use the tangent function, which is defined as \(\tan\theta=\frac{\text{opposite}}{\text{adjacent}}\) in a right - triangle. For \(\angle C = 30^{\circ}\), the opposite side to \(\angle C\) is \(AB\) and the adjacent side is \(AC\). So, \(\tan(30^{\circ})=\frac{AB}{AC}\).

Step2: Substitute values and solve for \(AB\)

We know that \(\tan(30^{\circ})=\frac{1}{\sqrt{3}}\) and \(AC = 6\). Substituting these values into the equation \(\tan(30^{\circ})=\frac{AB}{AC}\), we get \(\frac{1}{\sqrt{3}}=\frac{AB}{6}\). Cross - multiplying gives us \(AB=\frac{6}{\sqrt{3}}\). Rationalizing the denominator (multiplying numerator and denominator by \(\sqrt{3}\)), we have \(AB=\frac{6\sqrt{3}}{3}=2\sqrt{3}\)? Wait, no, wait. Wait, actually, we can also use the 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 side, the side opposite \(60^{\circ}\) is \(\sqrt{3}\) times the shortest side, and the hypotenuse is twice the shortest side. Wait, in triangle \(ABC\), \(\angle C = 30^{\circ}\), \(\angle A=90^{\circ}\), so \(\angle B = 60^{\circ}\). The side adjacent to \(30^{\circ}\) is \(AC = 6\) (adjacent to \(\angle C\)), the side opposite to \(30^{\circ}\) is \(AB\), and the hypotenuse is \(BC\). Wait, no, \(\tan\theta=\frac{\text{opposite}}{\text{adjacent}}\), so for \(\angle C\), opposite is \(AB\), adjacent is \(AC\). So \(\tan(30^{\circ})=\frac{AB}{AC}\), so \(AB = AC\times\tan(30^{\circ})\). But \(\tan(30^{\circ})=\frac{1}{\sqrt{3}}\), so \(AB = 6\times\frac{1}{\sqrt{3}}=\frac{6}{\sqrt{3}} = 2\sqrt{3}\)? Wait, no, that's wrong. Wait, no, actually, if we consider the tangent of \(60^{\circ}\), because \(\angle B = 60^{\circ}\), \(\tan(60^{\circ})=\frac{AC}{AB}\), because for \(\angle B\), opposite is \(AC = 6\), adjacent is \(AB\). So \(\tan(60^{\circ})=\sqrt{3}=\frac{6}{AB}\), so \(AB=\frac{6}{\sqrt{3}}=2\sqrt{3}\)? No, wait, \(\tan(60^{\circ})=\frac{\text{opposite}}{\text{adjacent}}\) for \(\angle B\), opposite is \(AC = 6\), adjacent is \(AB\), so \(\tan(60^{\circ})=\frac{AC}{AB}\), so \(\sqrt{3}=\frac{6}{AB}\), so \(AB=\frac{6}{\sqrt{3}} = 2\sqrt{3}\)? Wait, no, that's not right. Wait, I made a mistake. Let's start over.

In right - triangle \(ABC\), \(\angle A = 90^{\circ}\), \(\angle C = 30^{\circ}\), \(AC = 6\). We can use the cotangent function. \(\cot(30^{\circ})=\frac{\text{adjacent}}{\text{opposite}}=\frac{AC}{AB}\). Since \(\cot(30^{\circ})=\sqrt{3}\), then \(\sqrt{3}=\frac{6}{AB}\), so \(AB=\frac{6}{\sqrt{3}}=2\sqrt{3}\)? No, wait, no. Wait, \(\cot\theta=\frac{\text{adjacent}}{\text{opposite}}\), so for \(\angle C\), adjacent is \(AC = 6\), opposite is \(AB\), so \(\cot(30^{\circ})=\frac{AC}{AB}\), \(\cot(30^{\circ})=\sqrt{3}\), so \(\sqrt{3}=\frac{6}{AB}\), so \(AB=\frac{6}{\sqrt{3}} = 2\sqrt{3}\)? Wait, but that seems small. Wait, no, wait, maybe I mixed up the angles. Wait, \(\angle C = 30^{\circ}\), so the side opposite \(\angle C\) is \(AB\), the side adjacent is \(AC\). So \(\tan(30^{\circ})=\frac{AB}{AC}\), so \(AB = AC\times\tan(30^{\circ})=6\times\frac{1}{\sqrt{3}} = 2\sqrt{3}\approx3.464\). But wait, another way: in a 30 - 60 - 90 triangle, the sides are in the ratio \(x:x\sqrt{3}:2x\), where \(x\) is the side opposite \(30^{\circ}\), \(x\sqrt{3}\) is the side opposite \(60^{\circ}\), and \(2x\) is the hypotenuse. Here, the side opposite \(60^{\circ}\) is \(AC = 6\) (because \(\angle B = 60^{\circ}\), so the side opposite \(\angle B\) is \(AC\)). So if the side opposite \(60^{\circ}\) is \(x\sqrt{3}=6\), then \(x=\frac{6}{\sqrt{3}} = 2\sqrt{3}\), and the side opposite \(30^{\circ}\) (which is \(AB\)) is \(x = 2\sqrt{3}\)? Wait, no, that's correct. Wait, no, I think I confused the angles. Let's label the triangle: right - angle at \(A\), so \(AB\) and \(AC\) are the legs, \(BC\) is the hypotenuse. \(\angle C = 30^{\circ}\), so angle at \(C\) is \(30^{\circ}\), so the side opposite angle \(C\) is \(AB\), the side adjacent to angle \(C\) is \(AC\), and the hypotenuse is \(BC\). So in a 30 - 60 - 90 triangle, the side opposite \(30^{\circ}\) (AB) is the shortest side, the side opposite \(60^{\circ}\) (AC) is \(\sqrt{3}\) times the shortest side, and the hypotenuse (BC) is twice the shortest side. So if \(AC\) (opposite \(60^{\circ}\)) is \(6\), then \(AB\) (opposite \(30^{\circ}\)) is \(\frac{6}{\sqrt{3}}=2\sqrt{3}\). Yes, that's correct.

Answer:

\(2\sqrt{3}\)