1) (4√3)/2 m 2) (√3 + 3)/2 m 3) (3 - √3)/ 2 m 4) √3/2 m 5) √3 + 1m Solution: (3) (3 - √3) / 2 m From the figure, PQ is the tower and... View Article
1) √3 : 1 2) √3 : 2 3) 1 : 2 4) 2 : √3 Solution: (1) √3 : 1 In triangle EDA, tan 30o = ED / AD 1 / √3 = ED / AD = h / AD AD = h √3 In... View Article
1) ab 2) √ab 3) √(a/b) 4) √(b/a) Solution: (2) √ab In triangle ABC, tan θ = h / a h = a tan θ --- (i) In triangle ADB, tan (90 - θ) =... View Article
1) 100√3 m 2) 200√3/3 m 3) 100√3 /3 m 4) 200√3 m Solution: (2) 200√3/3 m BC = 100 / tan 60o DB = 100 / tan 30o DC = 100 √3 - (100 /... View Article
1) 30° 2) 45° 3) 60° 4) None of these Solution: (3) 60° ∠OPA = ∠OPC = π / 4 Let OP = n, OA = n, AC = 2n OP / OA = cos (π / 4) PA = √2 n... View Article
(1) sin (A + C) (2) sin (A - C) (3) b/R (4) none of these Solution: Given that s - a, s - b, s - c are in GP. So (s - b)2 = (s - a)(s - c)... View Article
(1) (r1 + r2)/(1 + cos C) (2) (r1 + r2)/(1 + sin C) (3) (r1 + r2)/(1 + tan C) (4) none of these Solution: We know r1 = ∆/(s - a) r2 =... View Article
(1) (1 - cos A)/(1 - cos B) (2) (1 + cos B)/(1 + cos A) (3) (1 - cos A)/(1 + cos B) (4) none of these Solution: a(a + c - b)/b(b + c - a) =... View Article
(1) k2 + 6k + 1 = 0 (2) k2 - 6k + 1 < 0 (3) k2 - 6k + 1 ≥ 0 (4) none of these Solution: Given tan B tan C = k We know in a triangle ABC,... View Article
(1) 3/2 (2) 2/3 (3) 1/5 (4) 5 Solution: Given 5 cos C + 6 cos B = 4 6 cos A + 4 cos C = 5 This is of the form b cos C + c cos B = a and c... View Article
1) 2 sin 320 2) 4 sin 320 3) 6 sin 320 4) 8 sin 320 Solution: Let A, B, C are the angles in AP. Then 2B = A + C Also A + B + C = 180 So B... View Article