Vertices of a triangle are (a,b/2) (a/2,b (0,0) and its area is 1/2 sq units then find relation between a and b
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Answer:
Given A(2,3,1),B(−2,2,0) and C(0,1,−1)
For △ABC,
a=BC=(−2−0)2+(2−1)2+(0+1)2
∴a=6
b=AC=(2−0)2+(3−1)2+(1+1)2
∴b=12
c=AB=(2+2)2+(3−2)2+(1)2
∴c=18
cos∠ABC=2aca2+c2−b2=2×32
×6
18+6−12=3
1
Step-by-step explanation:
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