An isosceles triangle ABC is right angled at B.D is a point inside the triangle ABC. P and Q are the feet of the perpendiculars drawn from D on the side AB and AC respectively of ∆ ABC. If AP = a cm, AQ = b cm and ∠BAD = 15°, sin 75°=

ABC is an isosceles triangle with AB = AC, A circle through B touching AC at the middle point of Q, intersects AB at P. Then AP : AB is : 

ABC is an isosceles triangle such that AB = AC and ∠B = 35°. AD is the median to the base BC. Then ∠BAD is

If ∆ABC is an isosceles triangle with ∠C = 90° and AC = 5 cm then AB is : 

In Δ ABC, D and E are points on AB and BC, respectively, such that DE is parallel to AC. If DE = 3 cm, AC = 5 cm and the area of trapezium ACED = 32 cm2, then what will be the area of Δ BDE?

ΔABC ~ ΔPQR. The perimeters of ΔABC and ΔPQR are 72 cm and 43.2 cm, respectively. What is the ratio of the are area of ΔABC to the area of ΔPQR?

The adjacent angles on a straight line are in the ratio 5 ∶ 4, then the smaller angle is –

Let ΔABC ∼ Δ PQR and ar(∆ABC)/ar(∆PQR) =4/25. If AB = 12 cm, BC = 8 cm and AC = 9 cm, then QR is equal to: