Question

In a AABC, 3 <A= 4<B = 6<C. Find all the angles of the triangle. (Hint: 3 <A = 4<B = 6<C = x 3 From the given figure, <A=x/3 and so on]​

Answer 1

ԨII :)

Step-by-step explanation:

• Given parameters

In ΔABC,

3∠A = 4∠B= 6∠C

• Let us consider x = 3∠A = 4∠B = 6∠C

• x = 3∠A

∠A = x/3 (1)

• x = 4∠B

∠B = x/4 (2)

• x = 6∠C

∠C = x/6 (3)

By using angle sum property

∠A + ∠B + ∠C = 1800

Put the values of ∠A, ∠B, ∠C

x/3 + x/4 + x/6 = 1800

Let us find the L.C.M of 3,4,6 i.e 12

(4x + 3x + 2x)/12 = 1800

9x = 2160

x = 2400

• Substitute the value of x in eqaution (1), (2) and (3)

∠A= x/3

∠A= 240/3 = 80°

∠B= x/4

∠B= 240/4= 60°

∠C= x/6

∠C= 240/6 = 40°

Answer 2

Step-by-step explanation:

Given parameters

In ΔABC,

3∠A = 4∠B= 6∠C

Let us consider x = 3∠A = 4∠B = 6∠C

x = 3∠A

∠A = x/3………………….(1)

x = 4∠B

∠B = x/4…………………..(2)

x = 6∠C

∠C = x/6…………………….(3)

By using angle sum property

∠A + ∠B + ∠C = 1800

Put the values of ∠A, ∠B, ∠C

x/3 + x/4 + x/6 = 1800

Let us find the L.C.M of 3,4,6 i.e 12

(4x + 3x + 2x)/12 = 1800

9x = 2160

x = 2400

Substitute the value of x in eqaution (1), (2) and (3)

∠A= x/3

∠A= 240/3 = 80°

∠B= x/4

∠B= 240/4= 60°

∠C= x/6

∠C= 240/6 = 40°