Quantcast

A community for students. Sign up today!

Here's the question you clicked on:

55 members online
  • 0 replying
  • 0 viewing

kumar69

  • one year ago

In triangle ABC, angleA=60 degree, prove that BC2=AB2+AC2+AB.AC

  • This Question is Open
  1. RolyPoly
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    Can you please double check the question? Is it "-AB.AC" instead of "+AB.AC" ?

  2. RadEn
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    use the cosine rule : |dw:1368950431568:dw| cos60 = (AB^2 + AC^2 - BC^2)/2ABAC

  3. RadEn
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    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

  4. RadEn
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    cross product out! AB * AC = AB^2 + AC^2 - BC^2

  5. RadEn
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    and actually, Rolypoly is right, that should "-AB.AC" not +AB AC i think there is a bit typo from u, @kumar69

  6. Not the answer you are looking for?
    Search for more explanations.

    • Attachments:

Ask your own question

Ask a Question
Find more explanations on OpenStudy

Your question is ready. Sign up for free to start getting answers.

spraguer (Moderator)
5 → View Detailed Profile

is replying to Can someone tell me what button the professor is hitting...

23

  • Teamwork 19 Teammate
  • Problem Solving 19 Hero
  • You have blocked this person.
  • ✔ You're a fan Checking fan status...

Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.

This is the testimonial you wrote.
You haven't written a testimonial for Owlfred.