Question

How do you find the cot of a 68 degree angle?

Answer 1

Hint: Here in this question, the cot function is given. Cot is a derived function from the basic trigonometric functions like tangent (tan). We can represent cot function in the form of sine and cosine as ${\displaystyle \frac{\cos \theta }{\sin \theta }}$. And, by using trigonometric ratios we have to find the cot of a 68-degree angle i.e. $\cot {68}^{\circ }$. So, we should know about the trigonometric ratios for different angles.

Let’s see some basic trigonometric functions:
$\Rightarrow$ Sine (sin)
$\Rightarrow$Cosine (cos)
$\Rightarrow$Tangent (tan)
When we say $\cot \theta$, here $\theta$ means angle in degrees.
Derived functions that are derived from basic trigonometric functions are:
$\Rightarrow$cosec$\theta$ = ${\displaystyle \frac{1}{\sin \theta }}$
$\Rightarrow$sec$\theta$ = ${\displaystyle \frac{1}{\cos \theta }}$
$\Rightarrow$tan$\theta$ = ${\displaystyle \frac{\sin \theta }{\cos \theta }}$ = ${\displaystyle \frac{1}{\cot \theta }}$
$\Rightarrow$cot$\theta$ = ${\displaystyle \frac{1}{\tan \theta }}$ = ${\displaystyle \frac{\cos \theta }{\sin \theta }}$
You know what cot$\theta$ is! Let’s find out.

So, from the figure,
$\Rightarrow$sin$\theta$ = ${\displaystyle \frac{perpendicular(P)}{hypotenuse(H)}}$
$\Rightarrow$cos$\theta$ = $${\displaystyle \frac{base(B)}{hypotenuse(H)}}$$
And we know that cot$\theta$ = ${\displaystyle \frac{\cos \theta }{\sin \theta }}$
So, cot$\theta$ = $${\displaystyle \frac{base(B)}{perpendicular(P)}}$$
Now, let’s have a look at some even and odd functions as well.
$\Rightarrow$sin(-x) = -sinx
$\Rightarrow$ cos(-x) = cosx
$\Rightarrow$ tan(-x) = -tanx
$\Rightarrow$ cot(-x) = -cotx
$\Rightarrow$ cosec(-x) = -cosecx
$\Rightarrow$sec(-x) = secx
Now, let’s make a table of trigonometric ratios for basic trigonometric functions i.e. sin, cos, tan, cot, sec, and cosec.

Trigonometric ratios(angle $\theta$ in degrees)	$0^{\circ }$	${30}^{\circ }$	${45}^{\circ }$	${60}^{\circ }$	${90}^{\circ }$
sin$\theta$	0	${\displaystyle \frac{1}{2}}$	${\displaystyle \frac{1}{\sqrt{2}}}$	${\displaystyle \frac{\sqrt{3}}{2}}$	1
cos$\theta$	1	${\displaystyle \frac{\sqrt{3}}{2}}$	${\displaystyle \frac{1}{\sqrt{2}}}$	${\displaystyle \frac{1}{2}}$	0
tan$\theta$	0	${\displaystyle \frac{1}{\sqrt{3}}}$	1	$\sqrt{3}$	$\mathrm{\infty }$
cosec$\theta$	$\mathrm{\infty }$	2	$\sqrt{2}$	${\displaystyle \frac{2}{\sqrt{3}}}$	1
sec$\theta$	1	${\displaystyle \frac{2}{\sqrt{3}}}$	$\sqrt{2}$	2	$\mathrm{\infty }$
cot$\theta$	$\mathrm{\infty }$	$\sqrt{3}$	1	${\displaystyle \frac{1}{\sqrt{3}}}$	0

We should also know what the graph of cotx looks like.

As we know that
cot$\theta$ = ${\displaystyle \frac{1}{\tan \theta }}$
So,
cot${68}^{\circ }$ = ${\displaystyle \frac{1}{\tan {68}^{\circ }}}$
So, cot${68}^{\circ }$ = $${\displaystyle \frac{1}{\tan {68}^{\circ }}}$$ = $${\displaystyle \frac{1}{2.475}}$$
$\therefore \cot {68}^{\circ }=$ 0.40

Note:
 You should remember all the functions and trigonometric ratios before solving any question related to trigonometry. As cot$\theta$ is a derived function, so you should also know the basic function of cot i.e. tan$\theta$ and how to derive the cot$\theta$. Find the exact value of the degree using the functions. Don’t keep the value up to 2 decimal places while calculating $\tan {68}^{\circ }$.