College Physics by Openstax Chapter 3 Problem 20

A new landowner has a triangular piece of flat land she wishes to fence. Starting at the west corner, she measures the first side to be 80.0 m long and the next to be 105 m. These sides are represented as displacement vectors A from B in Figure 3.59. She then correctly calculates the length and orientation of the third side C. What is her result?

Figure 3.59

Solution:

Consider the illustration shown.

We need to solve for an interior angle of the triangle. So, we need to solve for the value of α first. This can be done by simply subtracting the sum of 21 and 11 degrees from 90 degrees.

α=90∘−(21∘+11∘)α=58∘\begin{align*} \alpha & = 90 ^ \circ -\left( 21^\circ +11^\circ \right) \\ \alpha & = 58^\circ \end{align*}

Then, using the cosine law, we can now solve for the magnitude of vector C. That is

C2=A2+B2−2ABcos⁡αC2=(80 m)2+(105 m)2−2(80 m)(105 m)cos⁡58∘C2=8522.3564C=8522.3564C=92.3 m  (Answer)\begin{align*} C^2 & = A^2 + B^2 - 2AB \cos \alpha \\ C^2 & = \left( 80\ \text{m} \right)^2+\left( 105\ \text{m} \right)^2-2\left( 80\ \text{m} \right)\left( 105\ \text{m} \right) \cos 58^\circ \\ C^2 & = 8522.3564 \\ C & = \sqrt{8522.3564} \\ C & = 92.3 \ \text{m} \ \qquad \ {\color{DarkOrange} \left( \text{Answer} \right)} \end{align*}

Before we can solve for the value of θ, we need to know the value of β first. This can be done by using the sine law.

sin⁡β80 m=sin⁡58∘92.3 msin⁡β=(80 m)sin⁡58∘92.3 mβ=arcsin⁡[(80 m)sin⁡58∘92.3 m]β=47.3∘\begin{align*} \frac{\sin \beta}{80\ \text{m}} & = \frac{\sin 58^\circ }{92.3 \ \text{m}} \\ \sin \beta & = \frac{\left( 80 \ \text{m} \right)\sin 58^\circ }{92.3 \ \text{m}} \\ \beta & = \arcsin \left[ \frac{\left( 80 \ \text{m} \right)\sin 58^\circ }{92.3 \ \text{m}} \right] \\ \beta & = 47.3^\circ \end{align*}

Finally, we can solve for θ.

θ=(90∘+11∘)−βθ=(90∘+11∘)−47.3∘θ=53.7∘  (Answer)\begin{align*} \theta & = \left( 90 ^\circ +11^\circ \right) - \beta \\ \theta & = \left( 90 ^\circ +11^\circ \right) - 47.3^\circ \\ \theta & = 53.7^\circ \ \qquad \ {\color{DarkOrange} \left( \text{Answer} \right)} \end{align*}
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