Question : If $\cos(A+B)=0$ and $\sin(A-B)=0$, then what is the value of $\angle A$ and $\angle B$ ? ( $0^\circ < A, B \leq 90 ^\circ$)
Option 1: $20^\circ, 70^\circ$
Option 2: $45^\circ, 45^\circ$
Option 3: $60^\circ, 30^\circ$
Option 4: $15^\circ, 75^\circ$
New: SSC CHSL Tier 2 answer key released | SSC CHSL 2024 Notification PDF
Recommended: How to crack SSC CHSL | SSC CHSL exam guide
Don't Miss: Month-wise Current Affairs | Upcoming government exams
New: Unlock 10% OFF on PTE Academic. Use Code: 'C360SPL10'
Correct Answer: $45^\circ, 45^\circ$
Solution : Given: $\cos(A+B)=0$ and $\sin(A-B)=0$ The equation $\cos(A+B)=\cos 90^\circ$ $A+B=90^\circ$............(equation 1) The equation $\sin(A-B)=\sin 0^\circ$ $A-B=0^\circ$............(equation 2) Solving these two equations, we get: $\angle A = \angle B = 45^\circ$ Hence, the correct answer is $45^\circ, 45^\circ$.
Candidates can download this e-book to give a boost to thier preparation.
Result | Eligibility | Application | Admit Card | Answer Key | Preparation Tips | Cutoff
Question : If $0^{\circ} < \theta < 90^{\circ}$ and $2 \sin^{2}\theta +3\cos\theta =3$, then the value of $\theta$ is:
Question : If $\sin (\theta +18^{\circ})=\cos 60^{\circ}(0< \theta < 90^{\circ})$, then the value of $\cos 5\theta$ is:
Question : If $\cos A+\sin A=\sqrt{2}\cos A$, then $\cos A-\sin A$ is equal to: (where $0^{\circ}< A< 90^{\circ}$)
Question : If $\frac{\cos \theta}{(1+\sin \theta)}+\frac{\cos \theta}{(1-\sin \theta)}=4$ and $\theta$ is acute, then the value of $\theta$ is:
Question : In a $\triangle ABC$, if $2\angle A=3\angle B=6\angle C$, then the value of $\angle B$ is:
Regular exam updates, QnA, Predictors, College Applications & E-books now on your Mobile