Problems from 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. $$\cot (x)$$
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Development:

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. $$\cot (x)=\dfrac{4}{3}=1.333\ldots$$

Solution:

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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