Some Trigonometric Ratios and Formulae

0^{o} 
30^{o} 
45^{o} 
60^{o} 
90^{o} 
Sin 
0 
1/2 
1/âˆš2 
âˆš3/2 
1 
Cos 
1 
âˆš3/2 
1/âˆš2 
1/2 
0 
Tan 
0 
1/âˆš3 
1 
âˆš3 
âˆž 
sin(â€“Î¸) = â€“ sinÎ¸; cos(â€“Î¸) = cosÎ¸; tan(â€“Î¸) = â€“ tanÎ¸);
Example1
A car is being driven, in a straight line and at a uniform speed, towards the base of a vertical tower. The top of the tower is observed from the car and, in the process, it takes 10 minutes for the angle of elevation to change from 45Â° to 60Â°. After how much more time will this car reach the base of the tower?
 5()
 6( + )
 7(â€“ 1)
 8(â€“ 2)
Solution
Hence, x + d = xâˆš3,
So, x(âˆš3â€“1) = d
Car takes 10 mins in covering â€˜dâ€™ distance, so car will take 10/(âˆš3â€“1) min in covering â€˜xâ€™[x = d/(âˆš3â€“1)] distance.
Now, t = 10/(âˆš3 â€“ 1)
Multiplying both numerator and denominator by ( + 1), we get t = 5 ( + 1)
Example2
The angle of elevation of the top of a tower from a point 60m from its foot is 30^{0}. What is the height of the tower?
Solution
Let AB be the tower with its foot at A.
Let C be the point of observation.
Given âˆ ACB = 30^{0} and AC = 60m
Let C be the point of observation.
Given âˆ ACB = 30^{0} and AC = 60m
From right triangle BAC:
= tan 30^{0}
â‡’ AB = 60 tan 30^{0} = 60 = 20
= tan 30^{0}
â‡’ AB = 60 tan 30^{0} = 60 = 20
Example3
Two pillars of equal height stand on either side of a road. At a point on the road between the pillars the elevation of the tops of the pillars are 60^{0} and 30^{0}. Find the height of the pillars if it is given that the width of the road is 150m.
Solution
tan 60^{0} = = h/x
tan 30^{0} = 1/ = h/(150 â€“ X)
Solving we get: x = 37.5m, h = 64.95m.