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Some Trigonometric Ratios and Formulae

 

0o

30o

45o

60o

90o

Sin

0

1/2

1/√2

√3/2

1

Cos

1

√3/2

1/√2

1/2

0

Tan

0

1/√3

1

√3


sin(–θ) = – sinθ; cos(–θ) = cosθ; tan(–θ) = – tanθ);
 
cot(–θ) = – cotθ; sec(–θ) = secθ; cosec(–θ) = – cosecθ);
 
sin (900–θ) = cos θ; cos(900–θ) = sin θ
 
sin2θ = 2 sinθ.Cosθ
 
cos2θ = 2 cos2θ – 1 = 1 – 2 sin2θ
 
Description: 13339.png
 
Example-1
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 45° to 60°. After how much more time will this car reach the base of the tower?
  1. 5(Description: 10841.png)
  2. 6(Description: 10850.png + Description: 10856.png)
  3. 7(Description: 10863.png– 1)
  4. 8(Description: 10859.png– 2)
Solution
Description: 10870.png
Hence, x + d = x√3,
So, x(√3–1) = d
Car takes 10 mins in covering ‘d’ distance, so car will take 10/(√3–1) min in covering ‘x’[x = d/(√3–1)] distance.
Now, t = 10/(√3 – 1)
Multiplying both numerator and denominator by (Description: 10874.png + 1), we get t = 5 (Description: 10878.png + 1)

 
 
Example-2
The angle of elevation of the top of a tower from a point 60m from its foot is 300. 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 ∠ACB = 300 and AC = 60m
Description: 13364.png
From right triangle BAC:
Description: 10897.png = tan 300
⇒ AB = 60 tan 300 = 60Description: 10903.png = 20Description: 10910.png
 
 
Example-3
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 600 and 300. Find the height of the pillars if it is given that the width of the road is 150m.
Solution
tan 600 = Description: 10914.pngh/x
 
tan 300 = 1/Description: 10918.png = h/(150 – X)
 
Solving we get: x = 37.5m, h = 64.95m.
 




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