Question -
Answer -
Given:
In a тИЖABC, a = тИЪ2, b = тИЪ3 and c = тИЪ5┬а
By using the formulas,
We know, cos A = (b2┬а+ c2┬атАУa2)/2bc
By substituting the values we get,
= [(тИЪ3)2┬а+ (тИЪ5)2┬атАУ(тИЪ2)2] / [2 ├Ч тИЪ3 ├Ч тИЪ5]
= 3/тИЪ15
We know, Area of тИЖABC = 1/2 bc sin A
To find sin A:
Sin A = тИЪ(1 тАУ cos2┬аA) [by usingtrigonometric identity]
= тИЪ(1 тАУ (3/тИЪ15)2)
= тИЪ(1- (9/15))
= тИЪ(6/15)
Now,
Area of тИЖABC = 1/2 bc sin A
= 1/2 ├Ч тИЪ3 ├Ч тИЪ5 ├Ч тИЪ(6/15)
= 1/2 тИЪ6 sq. units
Hence proved.