Chapter 2

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

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3.

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

$\begin{array}{rcl} s & = & \frac{1}{2} (24.6 + 36.5 + 40.7) = 50.9 cm \\ A & = & \sqrt{50.9 (50.9 - 24.6) (50.9 - 36.5) (50.9 - 40.7)} \\ = & 443 {cm}^{2} \end{array}$ $\begin{array}{rcl} s & = & \frac{1}{2} (24.6 + 36.5 + 40.7) = 50.9 cm \\ A & = & \sqrt{50.9 (50.9 - 24.6) (50.9 - 36.5) (50.9 - 40.7)} \\ = & 443 {cm}^{2} \end{array}$
5. $\begin{array}{rcl} d^{2} & = & 125^{2} + 170 \\ d & = & \sqrt{125^{2} + 170^{2}} \\ = & 211 ft \end{array}$ $\begin{array}{rcl} d^{2} & = & 125^{2} + 170 \\ d & = & \sqrt{125^{2} + 170^{2}} \\ = & 211 ft \end{array}$
6. $\begin{array}{rcl} A & = & \frac{1}{2} h (b_{1} + b_{2}) \\ = & \frac{1}{2} (2.76) (9.96 + 4.70) = 20.2 m^{2} \end{array}$ $\begin{array}{rcl} A & = & \frac{1}{2} h (b_{1} + b_{2}) \\ = & \frac{1}{2} (2.76) (9.96 + 4.70) = 20.2 m^{2} \end{array}$
7. Let $m = mass of block$ $m = mass of block$
1. $m = 0.92 \times 10^{3} ({0.40}^{3}) = 59 kg$ $m = 0.92 \times 10^{3} ({0.40}^{3}) = 59 kg$
2. $A = 6 ({0.40}^{2}) = 0.96 m^{2}$ $A = 6 ({0.40}^{2}) = 0.96 m^{2}$
8. $\begin{array}{rclrcl} c & = & 2 π r & A & = & 4 π r^{2} \\ 21.0 & = & 2 π r & = & 4 π {(\frac{10.5}{π})}^{2} = \frac{4 ({10.5}^{2})}{π} \\ r & = & \frac{10.5}{π} & = & 140 {cm}^{2} \end{array}$ $\begin{array}{rclrcl} c & = & 2 π r & A & = & 4 π r^{2} \\ 21.0 & = & 2 π r & = & 4 π {(\frac{10.5}{π})}^{2} = \frac{4 ({10.5}^{2})}{π} \\ r & = & \frac{10.5}{π} & = & 140 {cm}^{2} \end{array}$
9. $\begin{array}{rcl} V & = & \frac{1}{3} π r^{2} h = \frac{1}{3} π ({2.08}^{2}) (1.78) \\ = & 8.06 m^{3} \end{array}$ $\begin{array}{rcl} V & = & \frac{1}{3} π r^{2} h = \frac{1}{3} π ({2.08}^{2}) (1.78) \\ = & 8.06 m^{3} \end{array}$
10. $\begin{array}{rcl} ∠ A C O + 64 ° & = & 90 ° & (tangent perpendicular to radius) \\ ∠ A C O & = & 26 ° \\ ∠ A = ∠ A C O & = & 26 ° & (isosceles triangle) \\ \frac{1}{2} \bar{C D} & = & 26 ° & (intercepted arc) \\ \bar{C D} & = & 52 ° \\ ∠ 1 & = & 52 ° & (central angle) \end{array}$ $\begin{array}{rcl} ∠ A C O + 64 ° & = & 90 ° & (tangent perpendicular to radius) \\ ∠ A C O & = & 26 ° \\ ∠ A = ∠ A C O & = & 26 ° & (isosceles triangle) \\ \frac{1}{2} \bar{C D} & = & 26 ° & (intercepted arc) \\ \bar{C D} & = & 52 ° \\ ∠ 1 & = & 52 ° & (central angle) \end{array}$

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11. $\begin{array}{rcl} ∠ C B O + ∠ 1 + 90 ° & = & 180 ° (sum of angles of triangle) \\ ∠ C B O + 52 ° + 90 ° & = & 180 ° \\ ∠ C B O & = & 180 ° - 52 ° - 90 ° = 38 ° \\ ∠ 2 + ∠ C B O & = & 180 ° (straight angle) \\ ∠ 2 + 38 ° & = & 180 ° \\ ∠ 2 & = & 142 ° \end{array}$ $\begin{array}{rcl} ∠ C B O + ∠ 1 + 90 ° & = & 180 ° (sum of angles of triangle) \\ ∠ C B O + 52 ° + 90 ° & = & 180 ° \\ ∠ C B O & = & 180 ° - 52 ° - 90 ° = 38 ° \\ ∠ 2 + ∠ C B O & = & 180 ° (straight angle) \\ ∠ 2 + 38 ° & = & 180 ° \\ ∠ 2 & = & 142 ° \end{array}$
12. $\begin{array}{rcl} r & = & \frac{1}{2} (2.25) cm \\ p & = & 3 (2.25) + \frac{1}{2} (2 π) [\frac{1}{2} (2.25)] \\ = & 10.3 cm \end{array}$ $\begin{array}{rcl} r & = & \frac{1}{2} (2.25) cm \\ p & = & 3 (2.25) + \frac{1}{2} (2 π) [\frac{1}{2} (2.25)] \\ = & 10.3 cm \end{array}$
13. $A = {2.25}^{2} - \frac{1}{2} π {[\frac{1}{2} (2.25)]}^{2} = 3.07 {cm}^{2}$ $A = {2.25}^{2} - \frac{1}{2} π {[\frac{1}{2} (2.25)]}^{2} = 3.07 {cm}^{2}$
14. $\begin{array}{rcl} A & = & \frac{1}{2} (50) [0 + 2 (90) + 2 (145) + 2 (260) \\ + 2 (205) + 2 (110) + 20] \\ = & 41, 000 {ft}^{2} \end{array}$ $\begin{array}{rcl} A & = & \frac{1}{2} (50) [0 + 2 (90) + 2 (145) + 2 (260) \\ + 2 (205) + 2 (110) + 20] \\ = & 41, 000 {ft}^{2} \end{array}$

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Table of Contents for Chapter 2

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Table of Contents for
Chapter 2