What is $\tan 60^{\circ}$?
A$\dfrac{1}{\sqrt{3}}$
B$1$
C$2$
D$\sqrt{3}$
Answer & Solution
Correct answer: D. $\sqrt{3}$
$\tan 60^{\circ} = \dfrac{\sin 60^{\circ}}{\cos 60^{\circ}} = \dfrac{\sqrt{3}/2}{1/2} = \sqrt{3}$.
Related questions
Penalty for wrong answers : THERE WILL BE PENALTY FOR WRONG ANSWERS MARKED BY A CANDIDATE Penalty for wrong answers : THERE WILL BE PENALTY FOR WRONG ANSWERS MARKED BY A CANDIDATE For [[1,2],[2,4]], the determinant equals:If A is invertible, |A^{−1}| equals:For a 3 × 3 non-singular A with |A| = 5, the value of |adj A| is:For a non-singular square matrix A of order n, |adj A| equals:For non-singular A, B of the same order, (AB)^{−1} equals:For a non-singular A, (A^{−1})^{−1} equals: