Quantcast

Got Homework?

Connect with other students for help. It's a free community.

  • across
    MIT Grad Student
    Online now
  • laura*
    Helped 1,000 students
    Online now
  • Hero
    College Math Guru
    Online now

Here's the question you clicked on:

55 members online
  • 0 replying
  • 0 viewing

AonZ Group Title

In the quadrilateral sketch opposite, <b and <D are right angles, AB= 3 BC = 3x and AD = 2DC = 2y. Use Pythagoras theorm and the cosine rule to show that <A = 45˚

  • one year ago
  • one year ago

  • This Question is Closed
  1. AonZ Group Title
    Best Response
    You've already chosen the best response.
    Medals 0

    |dw:1367567372412:dw|

    • one year ago
  2. RadEn Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

    |dw:1367568505062:dw|

    • one year ago
  3. RadEn Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

    first, look at triangle ABC and triangle ACD AC^2 (triangle ABC) = AC^2 (triangle ACD) (3x)^2 + x^2 = y^2 + (2y)^2 10x^2 = 5y^2 y^2 = 2x^2 ... (1) we knowed that : <A + <C = 180 so, <C = 180 - <A .... (2)

    • one year ago
  4. RadEn Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

    now, look at the triangle of BCD by using cosine rule, we have : BD^2 = BC^2 + CD^2 - 2 BC CD cos<C BD^2 = x^2 + y^2 - 2 xy cos<C ... (3) now, subtitute (1) and (2) to (3), giving us BD^2 = x^2 + 2x^2 - 2 xy cos(180 - A) BD^2 = 3x^2 - 2 xy (-cosA) BD^2 = 3x^2 + 2 xycosA

    • one year ago
  5. RadEn Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

    also, look at the triangle of ABD by using cosine rule, we have : BD^2 = AB^2 + AD^2 - 2 AB AD cosA BD^2 = (3x)^2 + (2y)^2 - 2 (3x) (2y) cosA BD^2 = 9x^2 + 4y^2 - 12xy cosA because y^2 = 2x^2 ---->y = x sqrt(2) then the equation above can be BD^2 = 9x^2 + 4(2x^2) -12x(x sqrt(2))cosA BD^2 = 9x^2 + 8x^2 -12x^2 sqrt(2)cosA BD^2 = 17x^2 - 12x^2 * sqrt(2)cosA

    • one year ago
  6. RadEn Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

    now, campare the values of both (BD^2) BD^2 = 3x^2 + 2 xycosA and BD^2 = 17x^2 - 12x^2 * sqrt(2)cosA it means : 17x^2 - 12x^2 * sqrt(2)cosA = 3x^2 + 2 xycosA combine the similar terms, giving us 17x^2 - 3x^2 = 2 xycosA + 12x^2 * sqrt(2)cosA 14x^2 = 2 xycosA + 12x^2 * sqrt(2)cosA

    • one year ago
  7. RadEn Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

    opps.. the value of y above must subtituted be x sqrt(2) so, we have : 14x^2 = 2 xycosA + 12x^2 * sqrt(2)cosA 14x^2 = 2 x(x sqrt(2))cosA + 12x^2 * sqrt(2)cosA 14x^2 = 2x^2 sqrt(2)cosA + 12x^2 * sqrt(2)cosA 14x^2 = 14x^2 sqrt(2) cosA divide by 14x^2 to both sides, becomes 1 = sqrt(2) cosA

    • one year ago
  8. RadEn Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

    then finally, 1 = sqrt(2) cosA or cosA = 1/sqrt(2) = 1/2 * sqrt(2) A = 45 degrees QED :)

    • one year ago
  9. AonZ Group Title
    Best Response
    You've already chosen the best response.
    Medals 0

    thanks so MUCH!!!

    • one year ago
  10. RadEn Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

    you're welcome :)

    • one year ago
    • Attachments:

See more questions >>>

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.