Inverse trigonometric ratios: cosecant, secant and cotangent

Given the triangle $ABC$, whose sides are $a = 3$, $b = 4$ and $c = 5$, being $x$ the angle of the $A$ vertex, compute the following values:

  1. $\sin (x)$
  2. $\cos (x)$
  3. $\tan (x)$
  4. $\csc (x)$
  5. $\sec (x)$
  6. $\cot (x)$

First, we define how long each side of the triangle is:

$$\overline{AB}=5 \qquad \overline{AC}=4 \qquad \overline{BC}=3$$

Then, once the length of every side is calculated, we proceed by computing the trigonometric ratios that have been told to:

  1. $\sin (x)=\dfrac{3}{5}=0.6$
  2. $\cos (x)=\dfrac{4}{5}=0.8$
  3. $\tan (x)=\dfrac{3}{4}=0.75$
  4. $\csc (x)=\dfrac{5}{3}=1.666\ldots$
  5. $\sec (x)=\dfrac{5}{4}=1.25$
  6. $\cot (x)=\dfrac{4}{3}=1.333\ldots$
  1. $\sin (x)=0.6$
  2. $\cos (x)=0.8$
  3. $\tan (x)=0.75$
  4. $\csc (x)=1.667$
  5. $\sec (x)=1.25$
  6. $\cot (x)=1.333$
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