If in ∆ABC, with usaly rotation,a=18, b=24, c=30 then sinA/2=?
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If in ∆ABC, with usaly rotation,a=18, b=24, c=30 then sinA/2= 1/√10
Step-by-step explanation:
Area of ∆ABC = (1/2) * b * c * sinA
Area of Triangle using Hero Formula
s = (a + b + b)/2 = ( 18 + 24 + 30)/2
=> 36
Area = √s(s-a)(s-b)(s-c)
= √36 * 18 * 12 * 6
= 216
(1/2) * b * c * sinA = 216
=> 24 * 30 * sinA = 432
=> sinA = 3/5
=> CosA = √1² - (3/5)² = 4/5
CosA = 1 - 2Sin²(A/2)
=> 2Sin²(A/2) = 1 - CosA
=> 2Sin²(A/2) = 1 - 4/5
=> 2Sin²(A/2) = 1/5
=> Sin²(A/2) = 1/10
=> Sin(A/2) = 1/√10
Another Method
using Cosine formula
CosA = (b² + c² - a²)/2bc
=> cosA = (24² + 30² - 18²)/(2 * 24 * 30)
=> CosA = 4/5
CosA = 1 - 2Sin²(A/2)
=> 2Sin²(A/2) = 1 - CosA
=> 2Sin²(A/2) = 1 - 4/5
=> 2Sin²(A/2) = 1/5
=> Sin²(A/2) = 1/10
=> Sin(A/2) = 1/√10
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