Question

Geometrically find the value of cos 30 degree

Answer 1

Dear Student,

Let us consider an Equilateral Trianlge $∆ABC$whose each side is of 2 Units. From Vertex C , draw and an altitude CD to opposite side AB which meets it at D. As per the properties of an Equilateral Triangle , the altitude bisects the angle $\angle C$ and the opposite side AB. Hence in Right Triangle $∆ADC$, $\angle ACD$= 30$°$ and AD = 1 unit.

​​​​​Let us now find out the Lenth of CD . In $∆ACD$ applying pythagorus theorum , we find that CD= $\sqrt{A{C}^2-A{D}^2}$ . Putting the values of CD , AC and AD in the above formula , we get CD= Sqrt ( 4-1) = $\sqrt{3}$
Now applying Cosine Formula in $∆ACD$ , Cos 30$°$ = Base / Hypotenuse , Where Base is CD = $\sqrt{3}$ and Hypotenuse is AC = 2.
Hence Cos 30$°$ = $\sqrt{3}$ $/$ 2. Hence Proved .

Regards