Question

The value of $\cos$ ${60}^{\circ }$ is

• A. 1/2
• B. √3
• C. √3/2
• D. 1/√2

Answer 1

The correct option is A

1/2

Consider an equilateral $\Delta ABC$ with sides of 2 units as shown in fig. Let there be a perpendicular dropped from A to side BC cuttng BC at D.
Now, in $\Delta ABDand\Delta ACD$:
AD is common
AB = AC (sides of an equilateral $\Delta ABC$)
and $\angle ADC=\angle ADB={90}^{\circ }\therefore \Delta ABD\cong \Delta ACD\left(RHScongruency\right)\Rightarrow \angle BAD=\angle CAD\left(correspondinganglesofcongruent\Delta s\right)\Rightarrow \angle BAD+\angle CAD=\angle BAC={60}^{\circ }\Rightarrow \angle BAD=\angle CAD={30}^{\circ }andBD=DC=\frac{1}{2}BC=1In\Delta ABD,A{B}^2=A{D}^2+B{D}^2\Rightarrow 2^2=1^2+B{D}^2\Rightarrow BD=\sqrt{3}cos\left(\angle BAD\right)=cos{30}^{\circ }=\frac{AD}{BA}=\frac{\sqrt{3}}{2}Also,cos\left(\angle ABD\right)=cos{60}^{\circ }=\frac{BD}{AB}=\frac{1}{2}$