Question

If 13 cos Θ = 12 and 180° < Θ < 360 calculate without the use of a calculator A) tan ΘB) (sin Θ + cos Θ)^2

Answer 1

$\begin{gathered} \text{Given} \\ 13\cos \theta=12 \end{gathered}$

First, solve the given in terms of cos Θ

$\begin{gathered} 13\cos \theta=12 \\ \frac{13\cos \theta}{13}=\frac{12}{13} \\ \\ \text{Therefore,} \\ \cos \theta=\frac{12}{13} \end{gathered}$

Next, draw a diagram to account for the fact that 180° < Θ < 360°

Solve for the opposite side using Pythagorean Theorem and make the resulting side negative (since it is below in the y-axis as shown on the diagram)

$\begin{gathered} a^2+b^2=c^2 \\ a^2+(12)^2=(13)^2 \\ a^2+144=169 \\ a^2=169-144 \\ a^2=25 \\ \sqrt[]{a^2}=\sqrt[]{25} \\ a=-5 \end{gathered}$

Part A:

Using the information above we can now solve for tan Θ

$\begin{gathered} \tan \theta=\frac{\text{opposite}}{\text{adjacent}} \\ \tan \theta=\frac{-5}{12} \\ \\ \tan \theta=-\frac{5}{12} \end{gathered}$

Part B:

$\begin{gathered} \sin \theta=\frac{\text{opposite}}{\text{hypotenuse}} \\ \sin \theta=\frac{-5}{13}=-\frac{5}{13} \\ \\ \cos \theta=\frac{12}{13} \\ \\ \mleft(\sin \theta+\cos \theta\mright)^2=​\Big(-\frac{5}{13}+\frac{12}{13}\Big)^2 \\ (\sin \theta+\cos \theta)^2=​\Big(\frac{7}{13}\Big)^2 \\ \\ (\sin \theta+\cos \theta)^2=\frac{49}{169} \end{gathered}$