1. The parallel sides of a trapezoid ABCD are in the ratio of 4 : 5. ABCD is divided into an isosceles triangle ABP and a parallelogram PBCD (as shown below). ABCD has a perimeter equal to 1120 meters and PBCD has a perimeter equal to 1000 meters. Find Sin∠ABC, given 2∠DAB = ∠BCD.
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By: anil on 05 May 2019 02.37 am
AB + BC + CD + AD = 1120 ------------Eqn(I) PB + BC + CD + PD = 1000 -------------Eqn(II) Subtracting eqn(II) from (I), we get : => AB - PB + (AD - PD) = 120 => AB - PB + AP = 120 => AB + AP = 120 + PB Now, if AB = PB, => AP = 120 => AD = 600 and BC = 480, then AB + PB + CD = 40, which is not possible (We know that BC = PD. If BC = PD = 480, then BC+PD = 960. PB + BC + CD + PD = 1000.
=> PB+CD = 40. Therefore, AB + PB+CD should be greater than 40). Similarly, AB = AP is also not possible. Thus $$AP = BP$$ => $$angle ABC = x + (180 - 2x) = (180 - x)$$ => $$sin angle ABC = sin (180 - x) = sin x$$ Also, perimeter of PBCD = $$10y = 1000$$ => $$y = 100$$ and perimeter of ABCD = $$AB + 10y = 1120$$ => $$AB = 120$$ Applying cosine rule in $$ riangle$$ ABP => $$cos x = frac{(AB)^2 + (AP)^2 - (BP)^2}{2 AB AP}$$ => $$cos x = frac{(120)^2 + (100)^2 - (100)^2}{2 imes 120 imes 100}$$ => $$cos x = frac{120}{200} = frac{3}{5}$$ $$ herefore sin x = sqrt{1 - (frac{3}{5})^2} = sqrt{1 - frac{9}{25}}$$ = $$sqrt{frac{16}{25}} = frac{4}{5}$$
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=> PB+CD = 40. Therefore, AB + PB+CD should be greater than 40). Similarly, AB = AP is also not possible. Thus $$AP = BP$$ => $$angle ABC = x + (180 - 2x) = (180 - x)$$ => $$sin angle ABC = sin (180 - x) = sin x$$ Also, perimeter of PBCD = $$10y = 1000$$ => $$y = 100$$ and perimeter of ABCD = $$AB + 10y = 1120$$ => $$AB = 120$$ Applying cosine rule in $$ riangle$$ ABP => $$cos x = frac{(AB)^2 + (AP)^2 - (BP)^2}{2 AB AP}$$ => $$cos x = frac{(120)^2 + (100)^2 - (100)^2}{2 imes 120 imes 100}$$ => $$cos x = frac{120}{200} = frac{3}{5}$$ $$ herefore sin x = sqrt{1 - (frac{3}{5})^2} = sqrt{1 - frac{9}{25}}$$ = $$sqrt{frac{16}{25}} = frac{4}{5}$$