Question
A 440 m long train crosses a platform twice its length
in 2 min. Find the speed of the train.Solution
Speed of train = (Length of train + Length of Platform)/Required time Length of train = 440 m Crossing time of platform = 2 × 60 = 120 sec Length of platform = 440 × 2 = 880 m Let the speed of train be x. Speed of train = (440 + 880)/120 ⇒ x = 11 m/s
If cos(90° − θ) = 3/5 for an acute angle θ, find:
Simplify:
15cos 41° cosec 49° - 6tan 55° tan 35°What is the value of [(sin x + sin y) (sin x – sin y)]/[(cosx + cosy) (cosy – cosx)]?
- If cos2B = sin(1.5B - 36 o ) , then find the measure of 'B'.
If 6sin²x + 2cos²x − 3 = 0, then find the value of sinx, given that 0° < x < 90°.
- Find the simplified value of the expression:sin 2 45 o  + sin 2 60 o  - (1/3) X tan 2 60 o
If x+  1/x = 2cosθ, then the value of x³+  1/x³ is
- If 4cos²A + 5sin²A = 4.5, then find the value of (sec²A - 1)

The Value of (sin38Ëš)/(cos52Ëš) + (cos12Ëš)/(sin78Ëš) - 4cos²60Ëš is