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}'$