Sarthaks Test
0 votes
39 views
in Arithmetic Progression by (7.9k points)
closed by

The angles of a quadrilateral are in A.P. whose common difference is 10°. Find the angles.

1 Answer

+1 vote
by (8.1k points)
selected by
 
Best answer

Let the angles be (a – 3d), (a – d), (a + d) and (a + 3d)

a – 3d + a – d + a + d + a + 3d = 360° (sum of angles of quadrilateral)

4a = 360°

a = 90°

According to question,

(a + d) – (a – d) = 10°

2d = 10°

d = 5°

Therefore,

(a – 3d) = 90° – 15° = 75°

(a – d) = 90° – 5° = 85°

(a + d) = 90° + 5° = 95°

(a + 3d) = 90° + 15° = 105°

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.

...