In a triangle ABC, AB = 7, AC = 8, BC = 3, what is cos(A)?
1 answer:
Use the cosine rule.
c=AB=7
b=AC=8
a=BC=3
a^2=b^2+c^2-2bc(cos(A))
=>
cos(A)=(b^2+c^2-a^2)/(2bc)
=(7^2+8^2-3^2)/(2*7*8)
=13/14
=0.9286
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