In a ∆ABC, a = √2, b = √3 and c = √5, if show that its area is 1/2 √6 sq. units.

Question

In a ∆ABC, a = √2, b = √3 and c = √5,  if show that its area is \(\cfrac12\sqrt6\) sq. units.

Answer 1

In any ΔABC we define ‘a’ as length of side opposite to ∠A , ‘b’ as length of side opposite to ∠B and ‘c’ as length of side opposite to ∠C .

Key point to solve the problem:

• Area of ΔABC = 0.5 x (product of any two sides) × ( sine of angle between them)

• Idea of cosine formula: Cos A = \(\cfrac{b^2+c^2-a^2}{2bc}\)

Given a = √2, b = √3 and c = √5

∵ Area of ΔABC = \(\cfrac12\) bc sin A

We need to find sin A 

As we know that - sin²A = 1 – cos²A {using trigonometric identity}