QUESTION IMAGE
Question
find the exact values of the six trigonometric ratios of the angle θ in the triangle.
sin(θ) =
cos(θ) =
tan(θ) =
csc(θ) =
sec(θ) =
cot(θ) =
Step1: Find the adjacent side
Let the adjacent side to angle $\theta$ be $a$, the opposite side be $b$ and the hypotenuse be $c$. Given $c = 6$ and $b=5$. Using the Pythagorean theorem $a=\sqrt{c^{2}-b^{2}}=\sqrt{6^{2}-5^{2}}=\sqrt{36 - 25}=\sqrt{11}$.
Step2: Calculate $\sin(\theta)$
By the definition of sine, $\sin(\theta)=\frac{b}{c}=\frac{5}{6}$.
Step3: Calculate $\cos(\theta)$
By the definition of cosine, $\cos(\theta)=\frac{a}{c}=\frac{\sqrt{11}}{6}$.
Step4: Calculate $\tan(\theta)$
By the definition of tangent, $\tan(\theta)=\frac{b}{a}=\frac{5}{\sqrt{11}}=\frac{5\sqrt{11}}{11}$.
Step5: Calculate $\csc(\theta)$
Since $\csc(\theta)=\frac{1}{\sin(\theta)}$, then $\csc(\theta)=\frac{6}{5}$.
Step6: Calculate $\sec(\theta)$
Since $\sec(\theta)=\frac{1}{\cos(\theta)}$, then $\sec(\theta)=\frac{6}{\sqrt{11}}=\frac{6\sqrt{11}}{11}$.
Step7: Calculate $\cot(\theta)$
Since $\cot(\theta)=\frac{1}{\tan(\theta)}$, then $\cot(\theta)=\frac{\sqrt{11}}{5}$.
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$\sin(\theta)=\frac{5}{6}$
$\cos(\theta)=\frac{\sqrt{11}}{6}$
$\tan(\theta)=\frac{5\sqrt{11}}{11}$
$\csc(\theta)=\frac{6}{5}$
$\sec(\theta)=\frac{6\sqrt{11}}{11}$
$\cot(\theta)=\frac{\sqrt{11}}{5}$