The angles of a triangle are in the ratio 1 : 2 : 7. Show that the ratio of the greatest side to the least sides is (√5 + 1) : (√5 – 1).

Let the angles of the triangle be x°, (2x)° and (7x)°.

Then x + 2x + 7x = 180°

⇒ x = 18°

So, the angles are A = 18°, B = 36° and C = 126°.

Clearly, the least side is 'a' and the greatest side is 'c'

From the sine rule,

Hence c : a = (√5 + 1) : (√5 – 1)

The sides of the triangle are 3 consecutive natural numbers and its largest angles is twice the smallest one. Find the sides of the triangle.

Let a = x – 1, b = x, c = x + 1

Given C = 2A

From sine rule,

(x² + 4x) 2(x – 1) = (x + 1)² (2x)

x(x + 4) (x – 1) = (x + 1)²(x)

x² – x + 4x – 4 = x² + 2x + 1

x² + 3x – 4 – x² – 2x – 1 = 0

x – 5 = 0

x = 5

∴ The sides of the triangle are 4, 5, 6.

If b = 4 cms, ∠A = 45°, ∠B = 30° then find 'a'

from sine rule

If a cos A = b cos B then prove that the triangle is either isosceles (or) right angled.

Given a cos A = b cos B

2R sin A cos A = 2R sin B cos B

sin 2A = sin 2B

⇒ sin 2A = sin 2B (or) sin 2A = sin(180 – 2B)

⇒ 2A = 2B (or) 2A = 180° – 2B

⇒ A = B (or) A + B = 90°

⇒ A = B (or) C = 90°

∴ The triangle is Isosceles (or) right angled.

Show that (b – a) cosC + c(cosB – cosA) =