Chapter 2

1. $\begin{array}{rcl} ∠ 1 + ∠ 3 + 90 ° & = & 180 ° (sum of angles of a triangle) \\ ∠ 3 & = & 52 ° (vertical angles) \\ ∠ 1 & = & 180 ° - 90 ° - 52 ° = 38 ° \end{array}$
2. $\begin{array}{rclc} ∠ 2 + ∠ 4 & = & 180 ° & (straight angle) \\ ∠ 4 & = & 52 ° & (corresponding angles) \\ ∠ 2 + 52 ° & = & 180 ° \\ ∠ 2 & = & 128 ° \end{array}$

1-1 Full Alternative Text
3.

$\frac{x}{8.0} = \frac{25.0}{10.0} x = \frac{8.0 (25.0)}{10.0} = 20.0 ft$
4. Use Hero’s formula:

$\begin{array}{rcl} s & = & \frac{1}{2} (24.6 + 36.5 + 40.7) = 50.9 cm \\ A & = & \sqrt{50.9 (50.9 - 24.6) (50.9 - 36.5) (50.9 - 40.7)} \\ = & 443 {cm}^{2} \end{array}$
5. $\begin{array}{rcl} d^{2} & = & 125^{2} + 170 \\ d & = & \sqrt{125^{2} + 170^{2}} \\ = & 211 ft \end{array}$
6. $\begin{array}{rcl} A & = & \frac{1}{2} h (b_{1} + b_{2}) \\ = & \frac{1}{2} (2.76) (9.96 + 4.70) = 20.2 m^{2} \end{array}$
7. Let $m = mass of block ...$

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Basic Technical Mathematics, 11th Edition by Allyn J. Washington, Richard Evans