Question

How to find the value of cos 30 geometrically?

Answer 1

Answer:

Step-by-step explanation:

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Answer 2

$\huge{Heya}$

$\small{ur /: answer:}$

$\bf{let's make an eq. ∆ ABC.}$

$\bf{Let each side of eq. ∆ ABC be 2a.}$

$\bf{Draw a perpendicular bisector 'D' on BC.}$

$\bf{It divides BC into two equal parts i.e BD = CD.}$

Also, BC = 1/2BD

=> 2a = 1/2BD

=> BD = a.

As,angle ADB = 90°

$\underline{bcoz D is perpendicular to BC.}$

and angle ABC = 60°

$\underline{bcoz the measure of each angle of an eq. ∆ = 60°.}$

Now, in right ∆ ABD :

By Applying Pythagoras theorem,

(AB)² = (BD)² + (AD)²

(2a)² = (BD)² + a²

=> (BD)² = 4a² - a²

=> (BD)² = 3a²

=> BD = √3a

now, cos 30° = AD/AB

$\underline{In this case, BD becomes perpendicular and AD becomes base.}$

=> cos 30° = √3a/2a

=> cos 30° = √3/2