Chapter 7 - Applications of Trigonometry and Vectors - Section 7.5 Applications of Vectors - 7.5 Exercises - Page 335: 5

The magnitude of the second force is 189.6 lb The magnitude of the resultant is 282.6 lb

Let $a = 176~lb$ Let $b$ be the second force. Let $c$ be the resultant force. When we use $a$ and $b$ to complete the parallelogram, the angles $A, B,$ and $C$ form a triangle, where $a$ is opposite angle $A$, $b$ is opposite angle $B$, and $c$ is opposite angle $C$. Angle $B = 41^{\circ}10'$. Angle $A = 78^{\circ}50'-41^{\circ}10'$ which is $A = 37^{\circ}40'$ We can use the law of sines to find $b$: $\frac{b}{sin~B} = \frac{a}{sin~A}$ $b = \frac{a~sin~B}{sin~A}$ $b = \frac{(176~lb)~sin~41^{\circ}10'}{sin~37^{\circ}40'}$ $b = 189.6~lb$ The magnitude of the second force is 189.6 lb Angle $C = 180^{\circ}- 78^{\circ}50' = 101^{\circ}10'$ We can use the law of sines to find $c$, which is the resultant: $\frac{c}{sin~C} = \frac{a}{sin~A}$ $c = \frac{a~sin~C}{sin~A}$ $c = \frac{(176~lb)~sin~101^{\circ}10'}{sin~37^{\circ}40'}$ $c = 282.6~lb$ The magnitude of the resultant is 282.6 lb