Chapter 9

1.
2. $\begin{array}{rcl} b^{2} & = & a^{2} + c^{2} - 2 a c \cos B \\ b & = & \sqrt{{22.5}^{2} + {30.9}^{2} - 2 (22.5) (30.9) \cos 78.6 °} \\ = & 34.4 \end{array}$ $\begin{array}{rcl} b^{2} & = & a^{2} + c^{2} - 2 a c \cos B \\ b & = & \sqrt{{22.5}^{2} + {30.9}^{2} - 2 (22.5) (30.9) \cos 78.6 °} \\ = & 34.4 \end{array}$
3. $\begin{array}{l} \begin{array}{rcl} x^{2} & = & {36.50}^{2} + {21.38}^{2} - 2 (36.50) (21.38) \cos 45.00 ° \\ x & = & 26.19 m \frac{21.38}{\sin α} = \frac{26.19}{\sin 45.00 °} \end{array} \\ \sin α = \frac{21.38 \sin 45.00 °}{26.19}, α = 35.26 ° \\ θ = 45.00 ° - 35.26 ° = 9.74 ° \end{array}$ $\begin{array}{l} \begin{array}{rcl} x^{2} & = & {36.50}^{2} + {21.38}^{2} - 2 (36.50) (21.38) \cos 45.00 ° \\ x & = & 26.19 m \frac{21.38}{\sin α} = \frac{26.19}{\sin 45.00 °} \end{array} \\ \sin α = \frac{21.38 \sin 45.00 °}{26.19}, α = 35.26 ° \\ θ = 45.00 ° - 35.26 ° = 9.74 ° \end{array}$

Displacement is 26.19 m, $9.74 °$ $9.74 °$ N of E.

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4. $\begin{array}{l} C = 180 ° - (18.9 ° + 104.2 °) = 56.9 ° \\ \frac{c}{\sin C} = \frac{a}{\sin A} \\ c = \frac{426 \sin 56.9 °}{\sin 18.9 °} = 1100 \end{array}$ $\begin{array}{l} C = 180 ° - (18.9 ° + 104.2 °) = 56.9 ° \\ \frac{c}{\sin C} = \frac{a}{\sin A} \\ c = \frac{426 \sin 56.9 °}{\sin 18.9 °} = 1100 \end{array}$
5. Since a is longest side, find A first.

$\begin{array}{l} a^{2} = b^{2} + c^{2} - 2 b c \cos A \\ \begin{array}{rcl} \cos A & = & \frac{b^{2} + c^{2} - a^{2}}{2 b c} = \frac{{3.29}^{2} + {8.44}^{2} - {9.84}^{2}}{2 (3.29) (8.44)} \\ A & = & 105.4 ° \\ \frac{b}{\sin B} & = & \frac{a}{\sin A} \\ \sin B & = & \frac{b \sin A}{a} = \frac{3.29 \sin 105.4 °}{9.84} \\ B & = & 18.8 ° \\ C & = & 180 ° - (105.4 ° + 18.8 °) = 55.8 ° \end{array} \end{array}$ $\begin{array}{l} a^{2} = b^{2} + c^{2} - 2 b c \cos A \\ \begin{array}{rcl} \cos A & = & \frac{b^{2} + c^{2} - a^{2}}{2 b c} = \frac{{3.29}^{2} + {8.44}^{2} - {9.84}^{2}}{2 (3.29) (8.44)} \\ A & = & 105.4 ° \\ \frac{b}{\sin B} & = & \frac{a}{\sin A} \\ \sin B & = & \frac{b \sin A}{a} = \frac{3.29 \sin 105.4 °}{9.84} \\ B & = & 18.8 ° \\ C & = & 180 ° - (105.4 ° + 18.8 °) = 55.8 ° \end{array} \end{array}$
6. $\begin{array}{l} \begin{array}{rcl} R_{x} & = & - 235, R_{y} = 152 \\ R & = & \sqrt{{(- 235)}^{2} + 152^{2}} = 280 \end{array} \\ \tan θ_{ref} = | \frac{152}{- 235} | = \frac{152}{235}, θ_{ref} = 32.9 ° \\ Quad . II : θ = 180 ° - 32.9 ° = 147.1 ° \end{array}$ $\begin{array}{l} \begin{array}{rcl} R_{x} & = & - 235, R_{y} = 152 \\ R & = & \sqrt{{(- 235)}^{2} + 152^{2}} = 280 \end{array} \\ \tan θ_{ref} = | \frac{152}{- 235} | = \frac{152}{235}, θ_{ref} = 32.9 ° \\ Quad . II : θ = 180 ° - 32.9 ° = 147.1 ° \end{array}$

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7. Let the force be F.

$\begin{array}{rcl} F_{x} & = & 871 \cos 284.3 ° = 215 kN \\ F_{y} & = & 871 \sin 284.3 ° = - 844 kN \end{array}$ $\begin{array}{rcl} F_{x} & = & 871 \cos 284.3 ° = 215 kN \\ F_{y} & = & 871 \sin 284.3 ° = - 844 kN \end{array}$

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8. $\begin{array}{l} \begin{array}{rclrcl} \frac{63.0}{\sin 148.5} & = & \frac{42.0}{\sin A} & \frac{x}{\sin 11.1 °} & = & \frac{63.0}{\sin 148.5 °} \\ \sin A & = & \frac{42.0 \sin 148.5 °}{63.0} & x & = & \frac{63.0 \sin 11.1 °}{\sin 148.5 °} \\ = & 23.2 mi \end{array} \\ \begin{array}{rcl} A & = & 20.4 ° \\ C & = & 180 ° - (148.5 ° + 20.4 °) = 11.1 ° \end{array} \end{array}$ $\begin{array}{l} \begin{array}{rclrcl} \frac{63.0}{\sin 148.5} & = & \frac{42.0}{\sin A} & \frac{x}{\sin 11.1 °} & = & \frac{63.0}{\sin 148.5 °} \\ \sin A & = & \frac{42.0 \sin 148.5 °}{63.0} & x & = & \frac{63.0 \sin 11.1 °}{\sin 148.5 °} \\ = & 23.2 mi \end{array} \\ \begin{array}{rcl} A & = & 20.4 ° \\ C & = & 180 ° - (148.5 ° + 20.4 °) = 11.1 ° \end{array} \end{array}$
9. $\begin{array}{l} \begin{array}{rcl} A_{x} & = & 449 \cos 74.2 ° B_{x} = 285 \cos 208.9 ° \\ A_{y} & = & 449 \sin 74.2 ° B_{y} = 285 \sin 208.9 ° \\ R_{x} & = & A_{x} + B_{x} = 449 \cos 74.2 ° + 285 \cos 208.9 ° \\ = & - 127.3 \\ R_{y} & = & A_{y} + B_{y} = 449 \sin 74.2 ° + 285 \sin 208.9 ° \\ = & 294.3 \\ R & = & \sqrt{{(- 127.3)}^{2} + {294.3}^{2}} = 321 \end{array} \\ \tan θ_{ref} = \frac{294.3}{127.3}, θ_{ref} = 66.6 °, θ = 113.4 ° \end{array}$ $\begin{array}{l} \begin{array}{rcl} A_{x} & = & 449 \cos 74.2 ° B_{x} = 285 \cos 208.9 ° \\ A_{y} & = & 449 \sin 74.2 ° B_{y} = 285 \sin 208.9 ° \\ R_{x} & = & A_{x} + B_{x} = 449 \cos 74.2 ° + 285 \cos 208.9 ° \\ = & - 127.3 \\ R_{y} & = & A_{y} + B_{y} = 449 \sin 74.2 ° + 285 \sin 208.9 ° \\ = & 294.3 \\ R & = & \sqrt{{(- 127.3)}^{2} + {294.3}^{2}} = 321 \end{array} \\ \tan θ_{ref} = \frac{294.3}{127.3}, θ_{ref} = 66.6 °, θ = 113.4 ° \end{array}$

$θ$ $θ$ is in second quadrant, since $R_{x}$ $R_{x}$ is negative and $R_{y}$ $R_{y}$ is positive.
10. $29.6 \sin 36.5 ° = 17.6$ $29.6 \sin 36.5 ° = 17.6$

$17.6 < 22.3 < 29.6$ $17.6 < 22.3 < 29.6$ means two solutions.

$\begin{array}{rcl} \frac{29.6}{\sin B} & = & \frac{22.3}{\sin 36.5 °}, \sin B = \frac{29.6 \sin 36.5 °}{22.3} \\ B_{1} & = & 52.1 ° C_{1} = 180 ° - 36.5 ° - 52.1 ° = 91.4 ° \\ B_{2} & = & 180 ° - 52.1 ° = 127.9 °, \\ C_{2} & = & 180 ° - 36.5 ° - 127.9 ° = 15.6 ° \end{array}$ $\begin{array}{rcl} \frac{29.6}{\sin B} & = & \frac{22.3}{\sin 36.5 °}, \sin B = \frac{29.6 \sin 36.5 °}{22.3} \\ B_{1} & = & 52.1 ° C_{1} = 180 ° - 36.5 ° - 52.1 ° = 91.4 ° \\ B_{2} & = & 180 ° - 52.1 ° = 127.9 °, \\ C_{2} & = & 180 ° - 36.5 ° - 127.9 ° = 15.6 ° \end{array}$

$\begin{array}{rclrcl} \frac{c_{1}}{\sin 91.4 °} & = & \frac{22.3}{\sin 36.5 °}, & c_{1} & = & \frac{22.3 \sin 91.4 °}{\sin 36.5 °} = 37.5 \\ \frac{c_{2}}{\sin 15.6 °} & = & \frac{22.3}{\sin 36.5 °}, & c_{2} & = & \frac{22.3 \sin 15.6 °}{\sin 36.5 °} = 10.1 \end{array}$ $\begin{array}{rclrcl} \frac{c_{1}}{\sin 91.4 °} & = & \frac{22.3}{\sin 36.5 °}, & c_{1} & = & \frac{22.3 \sin 91.4 °}{\sin 36.5 °} = 37.5 \\ \frac{c_{2}}{\sin 15.6 °} & = & \frac{22.3}{\sin 36.5 °}, & c_{2} & = & \frac{22.3 \sin 15.6 °}{\sin 36.5 °} = 10.1 \end{array}$

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11. Sum of x-components $= 0 :$ $= 0 :$

$\begin{array}{l} 185 \cos 305.6 ° + T_{2} \cos 90 ° + T_{1} \cos 221.7 ° = 0 \\ T_{1} = \frac{- 185 \cos 305.6 °}{\cos 221.7 °} = 144 N \end{array}$ $\begin{array}{l} 185 \cos 305.6 ° + T_{2} \cos 90 ° + T_{1} \cos 221.7 ° = 0 \\ T_{1} = \frac{- 185 \cos 305.6 °}{\cos 221.7 °} = 144 N \end{array}$

Sum of y-components $= 0 :$ $= 0 :$

$\begin{array}{l} 185 \sin 305.6 ° + T_{2} \sin 90 ° + T_{1} \sin 221.7 ° = 0 \\ 185 \sin 305.6 ° + T_{2} + (144.24) \sin 221.7 ° = 0 \\ \begin{array}{rcl} - T_{2} & = & - 246.38 \\ T_{2} & = & 246 N \end{array} \end{array}$ $\begin{array}{l} 185 \sin 305.6 ° + T_{2} \sin 90 ° + T_{1} \sin 221.7 ° = 0 \\ 185 \sin 305.6 ° + T_{2} + (144.24) \sin 221.7 ° = 0 \\ \begin{array}{rcl} - T_{2} & = & - 246.38 \\ T_{2} & = & 246 N \end{array} \end{array}$

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Table of Contents for Chapter 9

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Table of Contents for
Chapter 9