Special Right Triangles Exercise 2
Exercises
The leg opposite the 60° angle is \(7\sqrt{3}.\)
Solution Method:
For this problem, we have to draw our own picture. I know I have a right triangle, with a 60\(^\circ\) angle, and a hypotenuse of 14 inches. I need to find the length of the other two sides, the legs, so I’ll label them \(a\) and \(b\). Finally, I can calculate the measure of the third angle:\[180^{\circ}-90^{\circ}-60^{\circ}=30^{\circ}\]
I can now see this is a 30-60-90 triangle. A 30-60-90 triangle is a special triangle because it’s side lengths will always be in the ratio \(1:\sqrt{3}:2\). We can also write this ratio with a variable, \(x\), as \(x:x\sqrt{3}:2x\).
Using the side length we know, we can write this ratio for my triangle here. The longest side length of a triangle is always the hypotenuse, so \(14=2x\), so \(x=7\). Now we can rewrite the ratio \[7:7\sqrt{3}:14\] So we know that one side length is 7 in. and the other is \(7\sqrt{3}\) in. To determine which is \(a\) and which is \(b\), we need to recall the angle-side relationship of triangles: the largest side is opposite the largest angle, vice verse, and the shortest side is opposite the smallest angle, and vice versa. That means, because \(6^\circ > 30^\circ,\) the side opposite the 60\(^\circ\) angle is larger than the side opposite the 30\(^\circ\) angle, i.e. \(b>a\). So \(a=7\) and \(b=7\sqrt{3}\).
