Question

If the sides of a triangle abc satisfy the relation 4a^2+9b^2+c^2=2a(a+3b+c)

Answer 1

Step-by-step explanation:

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Answer 2

Your question is incomplete. Here is the full question:

If the sides of a triangle ABC satisfy the relation

4a²+9b²+c² = 2a(a+3b+c). Then find cos(A).

Concept:

Cosine rule: In a triangle ABC,

a² = b² + c² – 2bc·cos( ∠A), where a is the length of side opposite to ∠A,

b is the length of side opposite to ∠B, c is the length of side opposite to ∠C.

Given:

relation between sides ofΔABC

4a²+9b²+c² = 2a(a+3b+c).

Find:

The value of cos(A).

Solution:

given,

4a²+9b²+c² = 2a(a+3b+c)

⇒ 4a²+9b²+c² = 2a² + 6ab + 2ac

⇒ 2a²+ 9b²+c² - 6ab - 2ac = 0

⇒ (a²+ 9b² - 6ab) + (a²+ c²- 2ac) = 0

⇒ (a - 3b)² + (a - c)² = 0

As all a, b and c are real numbers, for the above equation to be satisfied,

a - 3b = 0     and   a - c = 0;

⇒ b = a/3;

⇒ c = a;

Using cosine rule,

a² = b² + c² – 2bc·cos( ∠A)

⇒ a² = (a/3)² + a² – 2(a²/3)·cos( ∠A)

⇒ a² = (a/3)² + a² – 2(a²/3)·cos( ∠A)

⇒ 2a² - a²/9 =  – 2(a²/3)·cos( ∠A)

⇒ 17a²/9 =  – 2(a²/3)·cos( ∠A)

⇒ cos( ∠A) = -(17/6)

∴ For the given relation, the cosine of ∠A will be cos( ∠A) = -(17/6).

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