Question : In a triangle $PQR$, $QR$ is produced to $S$. If $\angle PRS=(9x-15^{\circ}), \angle RPQ=2x$ and $\angle PQR = (4{x}+15^{\circ})$, what is the value of $x$?
Option 1: 55
Option 2: 20
Option 3: 10
Option 4: 75
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
Correct Answer: 10
Solution :
Given: $\angle PRS=(9x-15^{\circ}), \angle RPQ=2x$ and $\angle PQR=(4{x}+15^{\circ})$
Applying exterior angle property, we get:
$\angle PRS=\angle RPQ+\angle PQR$
⇒ $(9x-15^{\circ}) = 2x+(4{x}+15^{\circ})$
⇒ $3x=30^{\circ}$
⇒ $x=10^{\circ}$
Hence, the correct answer is 10.
Related Questions
Know More about
Staff Selection Commission Combined High ...
Result | Eligibility | Application | Admit Card | Answer Key | Preparation Tips | Cutoff
Get Updates BrochureYour Staff Selection Commission Combined Higher Secondary Level Exam brochure has been successfully mailed to your registered email id “”.