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
299 views
in Triangles by (36.4k points)
closed by

In an isosceles triangle, if the vertex angle is twice the sum of the base angles, calculate the angles of the triangle.

1 Answer

+1 vote
by (38.0k points)
selected by
 
Best answer

Let ΔABC be isosceles

Such that,

AB = BC

∠B = ∠C

Given, that vertex angle A is twice the sum of the base angles B and C.

i.e., ∠A = 2(∠B + ∠C)

∠A = 2(∠B + ∠B)

∠A = 2(2∠B)

∠A = 4∠B

Now,

We know that the sum of all angles of triangle = 180°

∠A + ∠B + ∠C = 180°

4∠B + ∠B + ∠B = 180°(Therefore, ∠A = 4∠B, ∠C = ∠B)

6∠B = 180°

∠B = \(\frac{180}{6}\)

= 30°

Since,

∠B = ∠C = 30°

And,

∠A = 4∠B

= 4 x 30° = 120°

Therefore, the angles of the triangle are 120°, 30°, 30°.

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

...