# Geometry applying Trigonometry

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\$

$\displaystyle \therefore 15\sin \theta -15\cos \theta =10$

$\displaystyle \therefore \sin \theta -\cos \theta =\frac{2}{3}$

$\displaystyle \therefore \frac{{\sqrt{2}}}{2}\sin \theta -\frac{{\sqrt{2}}}{2}\cos \theta =\frac{{\sqrt{2}}}{2}\times \frac{2}{3}$

$\displaystyle \therefore \sin \theta \cos 45{}^\circ -\cos \theta \sin 45{}^\circ =\frac{{\sqrt{2}}}{3}$

$\displaystyle \therefore \sin (\theta -45{}^\circ )=0.4714$

$\displaystyle \therefore \theta -45{}^\circ =28{}^\circ {8}'$

$\displaystyle \therefore \theta =73{}^\circ {8}'$