PROBLEMS ON TRIGONOMETRIC RATIOS

Problem 1 :

For the measures in the figure shown below, compute sine, cosine and tangent ratios of the angle θ.

Solution :

In the given right angled triangle, note that for the given angle θ, PR is the ‘opposite’ side and PQ is the ‘adjacent’ side.

Then, 

sinθ  =  opposite side/hypotenuse  =  PR/QR  =  35/37

cosθ  =  adjacent side/hypotenuse = PQ/QR  =  12/37

tanθ  =  opposite side / adjacent side  =  PR/PQ  =  35/12

Problem 2 :

Find the six trigonometric ratios of the angle θ using the diagram shown below. 

Solution :

In the given right angled triangle, note that for the given angle θ, AC is the ‘opposite’ side and AB is the ‘adjacent’ side.

And also, the length of the adjacent side 'AB' is not given. 

Find the length of AB.

By Pythagorean Theorem,

BC2  =  AB2 + AC2

252  =  AB2 + 72

625  =  AB2 + 49

Subtract 49 from each side. 

576  =  AB2

242  =  AB2

24  =  AB  

Then, 

sinθ  =  opposite side/hypotenuse  =   AC/ BC  =  7/25

cosθ  =  adjacent side/hypotenuse  =  AB/BC  =  24/25

tanθ  =  opposite side/adjacent side  =  AC/AB  =  7/24

cscθ  =  1/sinθ  =  25/7

secθ  =  1/cosθ  =  25/24

cotθ  =  1/tanθ  =  24/7

Problem 3 :

If tanA = 2/3, then find all the other trigonometric ratios. 

Solution :

tanA  =  opposite side/adjacent side  =  2/3

By Pythagorean Theorem,

AC2  =  AB2 + BC2

AC2  =  32 + 22

AC2  =  9 + 4

AC2  =  13

AC  =  √13

Then, 

sinA  =  opposite side/hypotenuse  =   BC/ AC  =  2/13

cosA  =  adjacent side/hypotenuse  =  AB/AC  =  3/√13

cscA  =  1/sinA  =  √13/2

secA  =  1/cosA  =  √13/3

cotA  =  1/tanA  =  3/2

Problem 4 :

If secθ  =  2/3, then find the value of

(2sinθ - 3cosθ)/(4sinθ - 9cosθ) 

Solution :

secθ  =  hypotenuse/adjacent side  =  13/5

By Pythagorean Theorem,

BC2  =  AB2 + AC2

132  =  52 + AC2

169  =  25 + AC2

Subtract 25 from each side. 

144  =  AC2

122  =  AC2

12  =  AC

Then, 

sinθ  =  opposite side/hypotenuse  =   AC/BC  =  12/13

cosθ  =  adjacent side/hypotenuse  =  AB/BC  =  5/13

(2sinθ - 3cosθ)/(4sinθ - 9cosθ) : 

=  (2 ⋅ 12/13 - 3 ⋅ 5/13)/(4 ⋅ 12/13 - 9 ⋅ 5/13)

=  (24/13 - 15/13)/(48/13 - 45/13)

=  [(24 - 15)/13]/[(48 - 45)/13]

=  (9/13)/(3/13)

=  (9/13)  (13/3)

=  9/3

=  3

So, 

(2sinθ - 3cosθ)/(4sinθ - 9cosθ)   =  3

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