the angle of the triangle are in AP if the greatest angle is twice the least find all the angles
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Answered by
622
Let three angles be a,a+d,a+2d
If they are in AP,
a+a+d+a+2d=180.
Therefore, a+d=60.
also, given that a+2d=2a
Hence, a=2d
Therefore, 3d=60 and d=20.
and a=60-d=60-20=40
therefore, 1st angle is a =40
2nd is a+d=40+20=60
3rd is a+2d=40+40=80
If they are in AP,
a+a+d+a+2d=180.
Therefore, a+d=60.
also, given that a+2d=2a
Hence, a=2d
Therefore, 3d=60 and d=20.
and a=60-d=60-20=40
therefore, 1st angle is a =40
2nd is a+d=40+20=60
3rd is a+2d=40+40=80
Answered by
294
Let the least angle be a and the next be a+d and the largest be a+2d
The series: a, a+d, a+2d
we know that the largest angle is twice the smallest
⇒ a+2d = 2a
⇒ 2d = 2a - a
⇒ 2d = a
⇒ d = a/2 -------------(1)
The series: a, a+a/2 , a+2a/2
= a, a+a/2 , a+a
These are the angles in a triangle
⇒ a+a+a/2 + a+a = 180
⇒ (2a+2a+a+2a+2a)/2 = 180
⇒ 9a = 180×2
⇒ 9a = 360
⇒ a = 360/9
⇒ a = 40
substitute a = 40 in (1)
d = a/2
d = 40/2
∴ d = 20
The series: a, a+d, a+2d
= 40, 40+20 , 40+(2×20)
= 40, 60, 80 are the angles of a triangle
The series: a, a+d, a+2d
we know that the largest angle is twice the smallest
⇒ a+2d = 2a
⇒ 2d = 2a - a
⇒ 2d = a
⇒ d = a/2 -------------(1)
The series: a, a+a/2 , a+2a/2
= a, a+a/2 , a+a
These are the angles in a triangle
⇒ a+a+a/2 + a+a = 180
⇒ (2a+2a+a+2a+2a)/2 = 180
⇒ 9a = 180×2
⇒ 9a = 360
⇒ a = 360/9
⇒ a = 40
substitute a = 40 in (1)
d = a/2
d = 40/2
∴ d = 20
The series: a, a+d, a+2d
= 40, 40+20 , 40+(2×20)
= 40, 60, 80 are the angles of a triangle
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