Use app×
Join Bloom Tuition
One on One Online Tuition
JEE MAIN 2025 Foundation Course
NEET 2025 Foundation Course
CLASS 12 FOUNDATION COURSE
CLASS 10 FOUNDATION COURSE
CLASS 9 FOUNDATION COURSE
CLASS 8 FOUNDATION COURSE
0 votes
776 views
in Polygons by (46.2k points)
closed by

In the diagram, CDP is a straight line, ΔAQD is equilateral ∠BAR= 90°, ∠QAR= 135°, ∠BCD = 106° and ∠ABC = 100°. Then, ∠PDQ equals.

(a) 39° 

(b) 21° 

(c) 41° 

(d) 53°

1 Answer

+1 vote
by (49.2k points)
selected by
 
Best answer

(c) 41°

ΔAQD is an equilateral Δ 

⇒ ∠QAD = ∠QDA = ∠AQD = 60° 

∠BAD = 360° – (135° + 90° + 60°) 

= 360° – 285° = 75° (Angles round a pt.) 

Also, in quad. ABCD, 

∠CDA = 360°– (100° + 106° + 75°) 

= 360° – 281° = 79°. 

∴ ∠PDQ =180° – (∠CDA+ ∠QDA)         (CDP is a st. line) 

= 180° – (79° + 60°) = 180° – 139° = 41°.

Welcome to Sarthaks eConnect: A unique platform where students can interact with teachers/experts/students to get solutions to their queries. Students (upto class 10+2) preparing for All Government Exams, CBSE Board Exam, ICSE Board Exam, State Board Exam, JEE (Mains+Advance) and NEET can ask questions from any subject and get quick answers by subject teachers/ experts/mentors/students.

Categories

...