The angles of a triangle are in AP. The greatest angle is twice the least. Find all the angles of the triangle.
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Answers
Answer:
The sum of all angles of the triangle = 180°.
(a - d) + a (a + d)=180°
a - d + a + a + d = 180°
3a = 180°
a = 60°
The greatest angle is twice the least:
= Greatest angle = 2 × least angle.
= a + d = 2(a - d)
= a + d = 2a - 2d
= d + 2d = 2a - a
3d = a
3d = 60°
d = 20°
a = 60°
=a - d
= 60° - 20°
= 40°
a + d
=60° + 20°
= 80°
The angles are 60° , 40° and 80°.
Given :
The angles of a triangle are in AP and the greatest angle is twice the least.
To Find :
All the angles of the triangle.
Solution :
Let the angles of a triangle be A, B and C.
Since, angles are in A.P, thus, A < B < C
By Angle sum property, we know;
A + B + C = 180° ------------ (i)
As the angles are in AP,
2B = A + C ---------- (ii)
Also, given, C = 2A ----------- (iii)
From (i) and (ii), we get
2B + B = 180° (As A + C = 2B)
∴ B = 60°
From (ii), we have
∴ 120° = A + C
⇒ 120° = A + 2A (As C = 2A)
∴ A = 40°
From (iii) we have, C = 2A
C = 2×40° = 80°
Hence, the angles are 40°, 60° and 80°.