Start learning 50% faster. Sign in now
Let the speed of train A be S1 and the speed of train B be S2. And length of train A be L1 and the length of train B be L2. According to the question, (S1 + S2) = (L1 + LB)/15 And, (S1 - S2) = (L1 + LB)/35 Here, the length of both the train is equal. So, (S1 + S2) × 15 = (S1 - S2) × 35 => 15 S1 + 15 S2 = 35 S1 – 35 S2 => 20 S1 = 50 S2 => S1/S2 = 50/20 = 5/2 Speed of slower train = n% of faster train => n% = (2/5) × 100 = 40% Therefore, (n × 2) = 40 × 2 = 80
sin2 9 ° + sin2 10 ° + sin2 11 ° + sin2 12 ° + ……… + sin2 81 ° = ?
...If 4cos2x - 3 = 0, and (0° > x > 90°), then find the value of 'x'.
If tan θ = (2/√5), then determine the value of cos2 θ
If √3cosec 2x = 2, then the value of x:
If x = (sin 30 ° + cos 30 ° )/sec 60 ° , then find the value of 4x.
If sin3A = cos(5A-30°), then what is the value of A? Given that 3A is an acute angle.
If sin x + cos x = √2 sin x, then the value of sin x - cos x is:
Simplify the given equation: