Question

mixed practice! find the value of each variable.
10.
11.
12.
13.

Explanation:

Step1: Solve problem 10

In a \(45 - 45-90\) triangle, the legs are equal and the hypotenuse \(c=\sqrt{2}a\) (where \(a\) is the length of a leg).
Given hypotenuse \(= 28\), let \(x = y\) (legs of \(45 - 45-90\) triangle).
Using \(c=\sqrt{2}a\), we have \(28=\sqrt{2}x\), then \(x=\frac{28}{\sqrt{2}}=\frac{28\sqrt{2}}{2}=14\sqrt{2}\). So \(x = 14\sqrt{2}\), \(y = 14\sqrt{2}\).

Step2: Solve problem 11

In a \(30 - 60-90\) triangle, if the side opposite \(30^{\circ}\) is \(y\), the side opposite \(60^{\circ}\) is \(y\sqrt{3}\) and the hypotenuse is \(2y\).
Given the side opposite \(60^{\circ}\) is \(16\), so \(y\sqrt{3}=16\), then \(y=\frac{16}{\sqrt{3}}=\frac{16\sqrt{3}}{3}\). And \(x = 2y=\frac{32\sqrt{3}}{3}\).

Step3: Solve problem 12

In a \(45 - 45-90\) triangle, legs are equal. So \(y = 4\), and \(x=\sqrt{4^{2}+4^{2}}=\sqrt{16 + 16}=\sqrt{32}=4\sqrt{2}\).

Answer:

1. \(x = 14\sqrt{2}\), \(y = 14\sqrt{2}\)
2. \(x=\frac{32\sqrt{3}}{3}\), \(y=\frac{16\sqrt{3}}{3}\)
3. \(x = 4\sqrt{2}\), \(y = 4\)