Question

solve any one of the 2 .......​

Answer 1

Step-by-step explanation:

(i) Given a : b : c = 4 : 5 : 6

Let a = 4k, b = 5k, c = 6k.

∴ cos C = (a² + b² - c²)/2ab

            = (16k² + 25k² - 36k²)/40k²

            = k²(16 + 25 - 36)/k²(40)

            = 5/40

            = 1/8.

∴ cos A = (b² + c² - a²)/2bc

             = (25k² + 36k² - 16k²)/60k²

             = 45k²/60k²

             = 3/4

We know that:

cos 2A = 2 cos² A - 1

            = 2 (3/4)² - 1

            = 2(9/16) - 1

            = 1/8

            = cos C

∴ cos 2A = cos C

∴ ∠C = 2∠A.

(ii)

(a)

By cosine law of triangles,

cosA = (b² + c² - a²)/2bc

cosB = (a² + c² - b²)/2ac

cosC = (a² + b² - c²)/2ab

Given: (cosA/a) + (cosB/b) + (cosC/c)

= (b² + c² - a²/2abc) + (c² + a² - b²/2abc) + (a² + b² - c²/2abc)

= (b² + c² - a² + c² + a² - b² + a² + b² - c²)/2abc

= (a² + b² + c²)/2abc.

Hope it helps!

Answer 2

Answer

OK, so here we go :) 

We'll have to use the cosine rule to be able to prove this. 

Draw a triangle with side lengths 4, 5 and 6 respectively. 

Angle A is opposite side 4, angle B is opposite the length 5, and angle C is opposite side length 6. 

Applying the cosine rule shows us : 

CosA = (b^2 + c^2 - a^2) /(2bc) 

CosA = (5^2 + 6^2 - 4^2) /(2×5×6) 

CosA = 3/4 

A = 41.41*.......... 

Let's do it for angle C, shall we? 

CosC = (a^2 + b^2 - c^2) /2ab 

CosC = (4^2 + 5^2 - 6^2) /(2×4×5) 

CosC = 1/8 

C = 82.82*............ 

Since 82.82 is twice the angle for A, 

we have proved that: C = 2 × ∠A