Question : If $\cos A=\frac{1}{2}, 0 \leq A \leq 90^{\circ}$, then what is the value of sin (180 - A)?
Option 1: $\frac{1}{2}$
Option 2: $\frac{\sqrt{3}}{2}$
Option 3: $\frac{1}{\sqrt{3}}$
Option 4: $1$
Correct Answer: $\frac{\sqrt{3}}{2}$
Solution : According to the question sin2A + cos2A = 1 ⇒ sin$^{2} A + \frac{1}{2}^{2}$ = 1 ⇒ sin$^{2} A + \frac{1}{4}$ = 1 ⇒ sin$^{2} A = 1 - \frac{1}{4}$ = $\frac{3}{4}$ ⇒ Sin A = $\frac{\sqrt{3}}{2}$ Now, ⇒ sin(180 − A) = sin A = $\frac{\sqrt{3}}{2}$ Hence, the correct answer is $\frac{\sqrt{3}}{2}$
Application | Eligibility | Selection Process | Result | Cutoff | Admit Card | Preparation Tips
Question : If $\cos A=\frac{15}{17}, 0 \leq A \leq 90^{\circ}$, then the value of $\cot(90° - A)$ is:
Option 1: $\frac{8}{15}$
Option 2: $\frac{2 \sqrt{2}}{15}$
Option 3: $\frac{\sqrt{2}}{15}$
Option 4: $\frac{7}{15}$
Question : If $(\cos \theta+\sin \theta):(\cos \theta-\sin \theta)=(\sqrt{3}+1):(\sqrt{3}-1), 0^{\circ}<\theta<90^{\circ}$, then what is the value of $\sec \theta$?
Option 1: 2
Option 2: $\sqrt{2}$
Option 3: 1
Option 4: $\frac{2 \sqrt{3}}{3}$
Question : If $\tan A=\frac{4}{3}, 0 \leq A \leq 90^{\circ}$, then find the value of $\sin A$.
Option 1: $\frac{3}{5}$
Option 2: $1$
Option 3: $\frac{3}{4}$
Option 4: $\frac{4}{5}$
Question : If $3+\cos ^2 \theta=3\left(\cot ^2 \theta+\sin ^2 \theta\right), 0^{\circ}<\theta<90^{\circ}$, then what is the value of $(\cos \theta+2 \sin \theta)$ ?
Option 1: $\frac{2 \sqrt{3}+1}{2}$
Option 2: $3 \sqrt{2}$
Option 3: $\frac{3 \sqrt{3}+1}{2}$
Option 4: $\frac{\sqrt{3}+2}{2}$
Question : If $\operatorname{cosec} A+\cot A=3$, $0 \leq A \leq 90^{\circ}$, then find the value of cos A.
Option 1: $\frac{3}{4}$
Option 2: $\frac{2}{5}$
Option 3: $\frac{3}{5}$
Regular exam updates, QnA, Predictors, College Applications & E-books now on your Mobile