If the lengths of the sides of a triangle are in AP and the greatest angle is double the smallest, then a ratio of lengths of the sides of this triangle is
- A.3 : 4 : 5
- B.4 : 5 : 6
- C.5 : 9 : 13
- D.5 : 6 : 7
Show answer and solution
Correct option: B
Solution
Let sides be a-d, a, a+d (AP), with smallest angle A opposite smallest side and greatest angle C = 2A opposite largest side.
By the sine rule, sin C = sin 2A = 2 sin A cos A.
Using the cosine rule: cos A = (a² + (a+d)² - (a-d)²) / (2·a·(a+d)) ... actually, let sides be b-d < b < b+d.
cos A = (b² + (b+d)² - (b-d)²) / (2·b·(b+d)) = (b² + 4bd) / (2b(b+d)) = (b + 4d) / (2(b+d))
sin C / sin A = (b+d)/(b-d) (sine rule ratio of sides opposite C and A)
sin 2A / sin A = 2 cos A = (b+d)/(b-d)
So: 2 · (b + 4d)/(2(b+d)) = (b+d)/(b-d)
(b + 4d)(b - d) = (b+d)²
b² + 3bd - 4d² = b² + 2bd + d²
bd = 5d², so b = 5d
Sides: b-d = 4d, b = 5d, b+d = 6d → ratio 4:5:6.
By the sine rule, sin C = sin 2A = 2 sin A cos A.
Using the cosine rule: cos A = (a² + (a+d)² - (a-d)²) / (2·a·(a+d)) ... actually, let sides be b-d < b < b+d.
cos A = (b² + (b+d)² - (b-d)²) / (2·b·(b+d)) = (b² + 4bd) / (2b(b+d)) = (b + 4d) / (2(b+d))
sin C / sin A = (b+d)/(b-d) (sine rule ratio of sides opposite C and A)
sin 2A / sin A = 2 cos A = (b+d)/(b-d)
So: 2 · (b + 4d)/(2(b+d)) = (b+d)/(b-d)
(b + 4d)(b - d) = (b+d)²
b² + 3bd - 4d² = b² + 2bd + d²
bd = 5d², so b = 5d
Sides: b-d = 4d, b = 5d, b+d = 6d → ratio 4:5:6.
[JEE Main 2019 · 8 Apr · Shift II]