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.
Explanation:
Let m∠DAB = θ ⇒ m∠BCD = 2θ □PBCD is a parallelogram. ∴ m∠DPB = 2θ m∠PBC = m∠PDC = (180 – 2θ) ∠DPB is an exterior angle of ∆PAB. ∴ By exterior angle theorem, m∠PBA = θ as, m∠DAB = θ Thus, in ∆PAB, PA = PB
∴ m∠ABC = θ + (180 – 2θ) = 180 – θ According to the given conditions, 10x + y = 1120 and 10x = 1000 Solving the two equations, we get x = 100 and y = 120 sin (180 – θ) = sin θ Applying sine rule to ∆PAB,
100sin θ=120sin (180-2θ)
100sin θ=120sin 2θ .....(i)
sin 2θ = 2 × sin θ × cos θ
∴ (i) becomes
100sin θ=1202×sin θ×cos θ
∴cos θ=35⇒sin θ=45
∴sin ∠ABC=sin (180-θ)=sin θ=45
Hence, option (a).
» Your doubt will be displayed only after approval.
Help us build a Free and Comprehensive Preparation portal for various competitive exams by providing us your valuable feedback about Apti4All and how it can be improved.