Question

FIND COSEC 30° AND COS 60° GEOMETRICALLY.PLEASE ANYONE ANSWER ME​

Answer 1

$\large\underline{\sf{Solution-}}$

Let us consider an equilateral triangle ABC with side 2a.

So, AB = BC = CA = 2a.

As triangle ABC is an equilateral triangle, so each angle of a triangle is 60°.

Construction :- From vertex A, draw AD perpendicular to BC.

As, we know, perpendicular drawn from one vertex to the opposite side is perpendicular bisector of opposite side and angle bisector of vertex.

So, AD is perpendicular bisector of BC.

It implies, BD = DC = a.

Now, in right angle triangle ABD

By using Pythagoras Theorem, we have

$\rm :\longmapsto\: {AB}^{2} = {BD}^{2} + {AD}^{2}$

On substituting the values of AB = 2a and BD = a, we get

$\rm :\longmapsto\: {(2a)}^{2} = {a}^{2} + {AD}^{2}$

$\rm :\longmapsto\: {4a}^{2} = {a}^{2} + {AD}^{2}$

$\rm :\longmapsto\: {4a}^{2} - {a}^{2} = {AD}^{2}$

$\rm :\longmapsto\: {3a}^{2} = {AD}^{2}$

$\bf\implies \:\: {AD} = \sqrt{3}a$

Now, Again in triangle ABD

$\rm :\longmapsto\:cos60\degree = \dfrac{BD}{AB}$

$\rm :\longmapsto\:cos60\degree = \dfrac{a}{2a}$

$\bf :\longmapsto\:\boxed{ \bf{ \: cos60\degree = \dfrac{1}{2} }}$

Again, in right triangle ABD

$\rm :\longmapsto\:cosec30\degree = \dfrac{AB}{BD}$

$\rm :\longmapsto\:cosec30\degree = \dfrac{2a}{a}$

$\rm :\longmapsto\:\boxed{ \bf{ \: cosec30\degree = 2 }}$

Additional Information:-

To find geometrically the value of sin60° and tan60°

In right triangle ABD

$\rm :\longmapsto\:sin60\degree = \dfrac{AD}{AB}$

$\rm :\longmapsto\:sin60\degree = \dfrac{ \sqrt{3}a }{2a}$

$\bf :\longmapsto\:sin60\degree = \dfrac{ \sqrt{3}}{2}$

Now,

$\rm :\longmapsto\:tan60\degree = \dfrac{AD}{BD}$

$\rm :\longmapsto\:tan60\degree = \dfrac{ \sqrt{3} a}{a}$

$\bf :\longmapsto\:tan60\degree = \sqrt{3}$