What is the value of $cos 60^{\circ}$ ?

1

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0

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2

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\frac{1}{2}

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Solution

The correct option is C $\frac{1}{2}$
Let $A B C$ be a right angled triangle wit right angle at $A$ , $∠ A B C = 60^{0}, ∠ B C A = 30^{0}$
Let $A B = a$
$\Rightarrow B C = 2 a$ , $A C = \sqrt{3} a$
Cosine of an angle is the ratio of adjacent side and the hypotenuse
$\Rightarrow c o s 60^{o} = \frac{A B}{B C} = \frac{a}{2 a} = \frac{1}{2}$

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Similar questions

c o s A = \sqrt{\frac{1 + c o s 2 A}{2}},

c o s 30^{0}

c o s 60^{0} = \frac{1}{2} .

cos A = \sqrt{\frac{1 + cos 2 A}{2}},

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cos 60^{\circ} = \frac{1}{2}

using the formula, sin A=whole root 1-cos A/2, find the value of sin30,it being given that cos60=1-2

c o s 60^{\circ} = \frac{1 - t a n^{2} 30^{\circ}}{1 + t a n^{2} 30^{\circ}} = \frac{1}{2}