Question : What is the value of $5 \sqrt{3} \cos 60^{\circ} \tan 30^{\circ}-3 \cos 0^{\circ}+3 \cos ^2 45^{\circ}+2 \sin ^2 60^{\circ}$?
Option 1: $2\sqrt{3}$
Option 2: $2.5$
Option 3: $3.5$
Option 4: $3\sqrt{2}$
Latest: SSC CGL preparation tips to crack the exam
Don't Miss: SSC CGL complete guide
New: Unlock 10% OFF on PTE Academic. Use Code: 'C360SPL10'
Correct Answer: $2.5$
Solution : $5 \sqrt{3} \cos 60^{\circ} \tan 30^{\circ}-3 \cos 0^{\circ}+3 \cos ^2 45^{\circ}+2 \sin ^2 60^{\circ}$ $= 5 \sqrt{3} \times \frac{1}{2} \times\frac{1}{\sqrt3}-3 \times1+3 \times \frac{1}{(\sqrt2)^2}+2(\frac{\sqrt3}{2})^2$ $= \frac{5}{2} -3+\frac{3}{2} + \frac{3}{2}$ $= \frac{5-6+3+3}{2}$ $= \frac{5}{2}$ $= 2.5$ Hence, the correct answer is $2.5$.
Candidates can download this ebook to know all about SSC CGL.
Answer Key | Eligibility | Application | Selection Process | Preparation Tips | Result | Admit Card
Question : What will be the value of $\frac{\sin 30^{\circ} \sin 40^{\circ} \sin 50^{\circ} \sin 60^{\circ}}{\cos 30^{\circ} \cos 40^{\circ} \cos 50^{\circ} \cos 60^{\circ}}$?
Question : What is the value of $\sin 30^{\circ}+\cos 30^{\circ}+\tan 30^{\circ}$?
Question : The value of $\left(\sin 30^{\circ} \cos 60^{\circ}-\cos 30^{\circ} \sin 60^{\circ}\right)$ is equal to:
Question : The value of $\frac{5 \cos ^2 60^{\circ}+4 \sec ^2 30^{\circ}-\tan ^2 45^{\circ}}{\tan ^2 60^{\circ}-\sin ^2 30^{\circ}-\cos ^2 45^{\circ}}$ is:
Question : The value of $\frac{2 \tan 60^{\circ}}{1+\tan ^2 60^{\circ}}$ is:
Regular exam updates, QnA, Predictors, College Applications & E-books now on your Mobile