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 sides AB and Ac respectively of △ABC. If AP = a cm, AQ = b cm and ∠BAD= 15 ∘ , sin 75∘ = 2b

Explanation

△ ABC is a right angled isosceles triangle right angled at B Here ∠A = ∠C 90∘ + 2∠A = 180∘ ∴ ∠A = ∠C = 45∘ ∘ (12) Given ∠BAD = 15∘ From △ ABC, ∠BAC = ∠BAD + ∠DAQ ⇒ 45∘ = 15∘ + ∠DAQ ∴ ∠DAQ = 30∘ From △DAQ, ∠AQD = 90∘ and ∠DAQ = 30∘ ∠AQD + ∠DAQ + ∠ADQ = 180∘ 90∘ + 30∘ + ∠ADQ = 180∘ ⇒ ∠ADQ = 60∘ F rom△ADQ, AQ sin 60∘ = AD 3 b 3 2 = AD ( ∵ sin 60∘ = 2 ) 2b AD = 3 In △AP D, ∠AP D = 90∘ and ∠P AD = 15∘ ∠AP D + ∠P AD + ∠ADP = 180∘ 90∘ + 15∘ + ∠ADP = 180∘ ⇒ ∠ADP = 75∘ From △ APD, AP sin 75∘ = AD 2b Substituting AD = 3 in above equation a 2b ⇒ sin 75∘ = ( 3 ) 3a ∴ sin 75∘ = 2b