anonymous
  • anonymous
In triangle ABC, angleA=60 degree, prove that BC2=AB2+AC2+AB.AC
Mathematics
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anonymous
  • anonymous
In triangle ABC, angleA=60 degree, prove that BC2=AB2+AC2+AB.AC
Mathematics
katieb
  • katieb
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At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

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anonymous
  • anonymous
Can you please double check the question? Is it "-AB.AC" instead of "+AB.AC" ?
RadEn
  • RadEn
use the cosine rule : |dw:1368950431568:dw| cos60 = (AB^2 + AC^2 - BC^2)/2ABAC
RadEn
  • RadEn
1/2 = (AB^2 + AC^2 - BC^2)/2*AB*AC multiplify by 2 to both side, giving us 1 = (AB^2 + AC^2 - BC^2)/AB*AC

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RadEn
  • RadEn
cross product out! AB * AC = AB^2 + AC^2 - BC^2
RadEn
  • RadEn
and actually, Rolypoly is right, that should "-AB.AC" not +AB AC i think there is a bit typo from u, @kumar69

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