Question

the angles of a triangle are in A. P.,the least being half the greatest. find the angles.
I'll mark the brainliest for the right ans

Answer 1

Let the angles be a, b and c

According to second condition,

a = c/2 ------(1)

Since they are in AP,

b - a = c - b

=> 2b = a + c

=> 2b = c/2 + c

=> 2b = 3c/2

=> b = 3c/4

Now,

By angle sum property of triangle,

a + b + c = 180

=> c/2 + 3c/4 + c = 180

=> ( 2c + 3c + 4c) / 4 = 180

=> 9c = 180 × 4

=> c = 20 × 4

=> c = 80°

First angle (a) = 80 /2 = 40°

Second angle (b) = 3 × 80 / 4 = 60°

Third angle (c) = 80°

Answer 2

Let the angles be a, b and c

According to second condition,

a = c/2 ------(1)

Since they are in AP,

b - a = c - b

=> 2b = a + c

=> 2b = c/2 + c

=> 2b = 3c/2

=> b = 3c/4

Now,

By angle sum property of triangle,

a + b + c = 180

=> c/2 + 3c/4 + c = 180

=> ( 2c + 3c + 4c) / 4 = 180

=> 9c = 180 × 4

=> c = 20 × 4

=> c = 80°

First angle (a) = 80 /2 = 40°

Second angle (b) = 3 × 80 / 4 = 60°

Third angle (c) = 80°

${ \tt{ \orange{hope \: it \: helped⚡}}}$