Question

The angles of a triangle are in A.P. If the greatest angle is twice the least, find all the angles of the triangle.


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Answer 1

Given:-

• All angles of ∆ are in A.P.
• Greatest angle = 2 × least angle.

To find:-

• All the angles of ∆

Solution:-

As angles are in A.P. then let the angles be (a - d), a & (a + d)

Here, least angle = (a - d)° & greatest angle = (a + d)°

As we know,

☛ Sum of 3 angles of ∆ = 180°

A/q,

→ (a + d) = 2(a - d)

→ a + d = 2a - 2d

→ a - 2a = -2d - d

→ -a = -3d

→ a = 3d ------ Equation (1)

Case 2:-

→ (a - d) + a + (a + d) = 180°

→ a - d + a + a + d = 180°

→ 3a = 180°

→ 3(3d) = 180°

→ 9d = 180°

→ d = 180/9

→ d = 20

So, commom difference (d) = 20

Now ,put this value in (Equation 1)

→ a = 3(20)

→ a = 60

Now all angles of ∆ : —

→ First angle = a - d

→ First angle = (60 - 20)°

→ First angle = 40°

$\rule{200}{1}$

→ Second angle = a

→ Second angle = 60°

$\rule{200}{1}$

→ Third angle = (a + d)

→ Third angle = (60 + 20)°

→ Third angle = 80°

Therefore,

All angles of triangle are : 40°,60° & 80° .

Answer 2

Answer:

hi mate.

Step-by-step explanation:

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 st equation.

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

= ( 2a + 2a + a + 2a + 2a ) / 2 = 180 .

= 9a = 180 * 2

= 9a = 360

= a = 360 / 9

= a = 40.

substitute a = 40 in ------ 1 st equation.

d = a / 2

d = 40 / 2

d = 20.

The series : a + ( a + d ) + ( a + 2d)

= 40 + ( 40 + 20 ) + ( 40 + ( 2 * 20 )

= 40 , 60 , 80 .

so, 40, 60, 80 are the three angles of the triangle.

hope it helps you .

please thank and mark my answer as brainliest.

sorry for late answering .