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?
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.
\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
\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.
\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.
\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*}
Advertisements
Advertisements