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 $\dfrac{\cos \theta }{\sin \theta }$. 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 $ = $\dfrac{1}{\sin \theta }$
$\Rightarrow $sec$\theta $ = $\dfrac{1}{\cos \theta }$
$\Rightarrow $tan$\theta $ = $\dfrac{\sin \theta }{\cos \theta }$ = $\dfrac{1}{\cot \theta }$
$\Rightarrow $cot$\theta $ = $\dfrac{1}{\tan \theta }$ = $\dfrac{\cos \theta }{\sin \theta }$
You know what cot$\theta $ is! Let’s find out.

So, from the figure,
$\Rightarrow $sin$\theta $ = $\dfrac{perpendicular(P)}{hypotenuse(H)}$
$\Rightarrow $cos$\theta $ = \[\dfrac{base(B)}{hypotenuse(H)}\]
And we know that cot$\theta $ = $\dfrac{\cos \theta }{\sin \theta }$
So, cot$\theta $ = \[\dfrac{base(B)}{perpendicular(P)}\]
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	$\dfrac{1}{2}$	$\dfrac{1}{\sqrt{2}}$	$\dfrac{\sqrt{3}}{2}$	1
cos$\theta $	1	$\dfrac{\sqrt{3}}{2}$	$\dfrac{1}{\sqrt{2}}$	$\dfrac{1}{2}$	0
tan$\theta $	0	$\dfrac{1}{\sqrt{3}}$	1	$\sqrt{3}$	$\infty $
cosec$\theta $	$\infty $	2	$\sqrt{2}$	$\dfrac{2}{\sqrt{3}}$	1
sec$\theta $	1	$\dfrac{2}{\sqrt{3}}$	$\sqrt{2}$	2	$\infty $
cot$\theta $	$\infty $	$\sqrt{3}$	1	$\dfrac{1}{\sqrt{3}}$	0

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

As we know that
cot$\theta $ = $\dfrac{1}{\tan \theta }$
So,
cot${{68}^{\circ }}$ = $\dfrac{1}{\tan {{68}^{\circ }}}$
So, cot${{68}^{\circ }}$ = \[\dfrac{1}{\tan {{68}^{\circ }}}\] = \[\dfrac{1}{2.475}\]
$\therefore \cot {{68}^{\circ }}=$ 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 $\tan {{68}^{\circ }}$.