
Find the value of $\cos ec$${{30}^{\circ }}$ geometrically?
Answer
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Hint: Before starting the question, it is very important to learn all the definitions of all the $6$ trigonometric ratios which are $\sin ,\cos ,\tan ,\sec ,\cos ec,\cot $. To find the value of $\cos ec$${{30}^{\circ }}$ geometrically, we need to make us of an equilateral triangle. We all know that an equilateral triangle is a type of triangle in which all sides are equal. We drop a perpendicular from one vertex to it’s opposite side. We make use of some properties of the triangle and then find the value of $\cos ec$${{30}^{\circ }}$.
Complete step by step solution:
Picture for reference :
Let us take a triangle $ABC$ which is an equilateral triangle.
Let the length of the side be $a$.
Now let us drop a perpendicular named $AD$ from vertex $A$ to the opposite side $BC$.
We know that in an equilateral triangle the medians, perpendiculars and angular bisectors over-lap and are the same. So our perpendicular $AD$, bisects the side $BC$. And the mid-point is $D$. The perpendicular also bisects the apex angle which is $A$.
$\angle BAD=\angle DAC=\dfrac{{{60}^{\circ }}}{2}={{30}^{\circ }}$
$BD=DC=\dfrac{a}{2}$.
We know that the $\angle ADB={{90}^{\circ }}$.
We know that the definition of $\cos ec\theta $.
It is the following :
$\cos ec\theta =\dfrac{hypotenuse}{opposite}$.
And in this case our $\theta $ is ${{30}^{\circ }}$ .
From the $\Delta BAD$ ,
$\cos ec{{30}^{\circ }}=\dfrac{hypotenuse}{opposite}=\dfrac{AB}{BD}=\dfrac{a}{\left( \dfrac{a}{2} \right)}=2$ .
$\therefore $ Hence, the value of $\cos ec$${{30}^{\circ }}$ geometrically is $2$.
Note: We should remember all the definitions of all the $6$ trigonometric ratios so as to solve the questions quickly. If not $6$, we should be well-versed with definitions and all the values of $\sin ,\cos ,\tan $. If we inverse these $3$, we would get $co\sec ,\sec ,\cot $ respectively. We should be very careful doing calculations as it may lead to wrong results. It is very important to know all the other trigonometric identities and formulae.
Complete step by step solution:
Picture for reference :
Let us take a triangle $ABC$ which is an equilateral triangle.
Let the length of the side be $a$.
Now let us drop a perpendicular named $AD$ from vertex $A$ to the opposite side $BC$.
We know that in an equilateral triangle the medians, perpendiculars and angular bisectors over-lap and are the same. So our perpendicular $AD$, bisects the side $BC$. And the mid-point is $D$. The perpendicular also bisects the apex angle which is $A$.
$\angle BAD=\angle DAC=\dfrac{{{60}^{\circ }}}{2}={{30}^{\circ }}$
$BD=DC=\dfrac{a}{2}$.
We know that the $\angle ADB={{90}^{\circ }}$.
We know that the definition of $\cos ec\theta $.
It is the following :
$\cos ec\theta =\dfrac{hypotenuse}{opposite}$.
And in this case our $\theta $ is ${{30}^{\circ }}$ .
From the $\Delta BAD$ ,
$\cos ec{{30}^{\circ }}=\dfrac{hypotenuse}{opposite}=\dfrac{AB}{BD}=\dfrac{a}{\left( \dfrac{a}{2} \right)}=2$ .
$\therefore $ Hence, the value of $\cos ec$${{30}^{\circ }}$ geometrically is $2$.
Note: We should remember all the definitions of all the $6$ trigonometric ratios so as to solve the questions quickly. If not $6$, we should be well-versed with definitions and all the values of $\sin ,\cos ,\tan $. If we inverse these $3$, we would get $co\sec ,\sec ,\cot $ respectively. We should be very careful doing calculations as it may lead to wrong results. It is very important to know all the other trigonometric identities and formulae.
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