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AonZ
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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
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

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AonZ Group TitleBest ResponseYou've already chosen the best response.0
dw:1367567372412:dw
 one year ago

RadEn Group TitleBest ResponseYou've already chosen the best response.1
dw:1367568505062:dw
 one year ago

RadEn Group TitleBest ResponseYou've already chosen the best response.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

RadEn Group TitleBest ResponseYou've already chosen the best response.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

RadEn Group TitleBest ResponseYou've already chosen the best response.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

RadEn Group TitleBest ResponseYou've already chosen the best response.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

RadEn Group TitleBest ResponseYou've already chosen the best response.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

RadEn Group TitleBest ResponseYou've already chosen the best response.1
then finally, 1 = sqrt(2) cosA or cosA = 1/sqrt(2) = 1/2 * sqrt(2) A = 45 degrees QED :)
 one year ago

AonZ Group TitleBest ResponseYou've already chosen the best response.0
thanks so MUCH!!!
 one year ago

RadEn Group TitleBest ResponseYou've already chosen the best response.1
you're welcome :)
 one year ago
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