Chapter 2 - Acute Angles and Right Triangles - Section 2.1 Trigonometric Functions of Acute Angles - 2.1 Exercises - Page 55: 71

a) $60^{\circ}$ b) $k$ c) $\sqrt 3$ $k$ d) In a $30^{\circ}$ - $60^{\circ}$ right triangle, the hypotenuse is always 2 times as long as the shorter leg, and the longer leg has a length that is $\sqrt 3$ times as long as that of the shorter leg. Also, the shorter leg is opposite the $30^{\circ}$ angle, and the longer leg is opposite the $60^{\circ}$ angle.

a) Since it is an equilateral triangle, this means all angles measure the same. We will name the angle $\theta$. Remember the sum of the interior angles in an equilateral triangle is equal to $180^{\circ}$. So, $\theta+\theta+\theta=180$ $3\theta=180^{\circ}$ $\theta=\frac{180^{\circ}}{3}$ $\theta=60^{\circ}$ b) Dividing the equilateral triangle creates two $30^{\circ} - 60^{\circ}$ right triangles. Let's name the side opposite of the $30^{\circ}$ angle: $x$. We can use the trigonometric function for sine to solve for x, where $2k$ is the hypotenuse and $x$ is the opposite side. $\sin(30^{\circ})=\frac{x}{2k}$ $\frac{1}{2}=\frac{x}{2k}$ $x=\frac{2k}{2}$ $x=k$ c) Let's call the perpendicular length from angle A: $p$. This would be the adjacent side to the $30^{\circ}$ angle. We can use the trigonometric function for cosine to solve for p, where $2k$ is the hypotenuse. $\cos(30^{\circ})=\frac{p}{2k}$ $\frac{\sqrt 3}{2}=\frac{p}{2k}$ $p=\frac{\sqrt 3 (2k) }{2}$ $p=\sqrt 3$ $k$ d) Since $x=k$ is the shorter leg and the hypotenuse is $2k$, the hypotenuse is always 2 times as long as the shorter leg. The longer length is $p$ and it's $\sqrt 3$ $k$, so the longer length is $\sqrt 3$ times as long as the shorter leg. From the description of the problem we know the $30^{\circ}$ is opposite the shorter leg $k$. Since it is a $30^{\circ} - 60^{\circ}$ right triangle, the longer leg has to be opposite $60^{\circ}$ angle.