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AonZ

  • 3 years ago

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˚

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  1. AonZ
    • 3 years ago
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    |dw:1367567372412:dw|

  2. RadEn
    • 3 years ago
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    |dw:1367568505062:dw|

  3. RadEn
    • 3 years ago
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    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)

  4. RadEn
    • 3 years ago
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    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

  5. RadEn
    • 3 years ago
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    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

  6. RadEn
    • 3 years ago
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    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

  7. RadEn
    • 3 years ago
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    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

  8. RadEn
    • 3 years ago
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    then finally, 1 = sqrt(2) cosA or cosA = 1/sqrt(2) = 1/2 * sqrt(2) A = 45 degrees QED :)

  9. AonZ
    • 3 years ago
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    thanks so MUCH!!!

  10. RadEn
    • 3 years ago
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    you're welcome :)

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