The trapezium ABCD is right angled at A and at D. and AB is parallel to DC. ∠ABC = θ, AB = l0 cm and BC = 15 cm.
(i) Express AD and DC in terms of θ.
(ii) Hence find the the value of θ for which its perimeter is 45 cm.
Solution
Draw CE ⊥ AB.
Let AD = x and CD = y, then CE = x, AE = y and BE = 10 - y.
Since $ \displaystyle \frac{{CE}}{{BC}}=\sin \theta, $
$ \displaystyle CE=BC\sin \theta $
$ \displaystyle x=15\sin \theta $
Similarly, $ \displaystyle \frac{{BE}}{{BC}}=\cos \theta ,$
$ \displaystyle BE=BC\cos \theta $
$ \displaystyle 10-y=15\cos \theta $
$ \displaystyle y=10-15\cos \theta $
By the problem, perimeter of ABCD = 45 cm
$ \displaystyle \therefore 10+15+x+y=45$
$ \displaystyle \therefore 10+15+15\sin \theta +10-15\cos \theta =45\ $
$ \displaystyle \therefore 15\sin \theta -15\cos \theta =10$
$ \displaystyle \therefore \sin \theta -\cos \theta =\frac{2}{3}$
$ \displaystyle \therefore \frac{{\sqrt{2}}}{2}\sin \theta -\frac{{\sqrt{2}}}{2}\cos \theta =\frac{{\sqrt{2}}}{2}\times \frac{2}{3}$
$ \displaystyle \therefore \sin \theta \cos 45{}^\circ -\cos \theta \sin 45{}^\circ =\frac{{\sqrt{2}}}{3}$
$ \displaystyle \therefore \sin (\theta -45{}^\circ )=0.4714$
$ \displaystyle \therefore \theta -45{}^\circ =28{}^\circ {8}'$
$ \displaystyle \therefore \theta =73{}^\circ {8}'$
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