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, 

sin θ  =  opposite side / hypotenuse  =  PR/QR  =  35/37

cos θ  =  adjacent side / hypotenuse  =  PQ/QR  =  12/37

tan θ  =  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, 

sin θ  =  opposite side / hypotenuse  =   AC/ BC  =  7/25

cos θ  =  adjacent side / hypotenuse  =  AB/BC  =  24/25

tan θ  =  opposite side / adjacent side  =  AC/AB  =  7/24

csc θ  =  1 / sin θ  =  25/7

sec θ  =  1 / cos θ  =  25/24

cot θ  =  1 / tan θ  =  24/7

Problem 3 :

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

Solution : 

tan A  =  opposite side / adjacent side  =  2 / 3

By Pythagorean Theorem,

AC²  =  AB² + BC²

AC²  =  3² + 2²

AC²  =  9 + 4

AC²  =  13

AC  =  √13

Then, 

sin A  =  opposite side / hypotenuse  =   BC/ AC  =  2/√13

cos A  =  adjacent side / hypotenuse  =  AB/AC  =  3/√13

csc A  =  1 / sin A  =  √13/2

sec A  =  1 / cos A  =  √13/3

cot A  =  1 / tan A  =  3/2

Problem 4 :

If sec θ  =  2/3, then find the value of

(2sin θ - 3cos θ) / (4sin θ - 9cos θ) 

Solution : 

sec θ  =  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, 

sin θ  =  opposite side / hypotenuse  =   AC/BC  =  12/13

cos θ  =  adjacent side / hypotenuse  =  AB/BC  =  5/13

(2sin θ - 3cos θ) / (4sin θ - 9cos θ) : 

=  (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, 

(2sin θ - 3cos θ) / (4sin θ - 9cos θ)   =  3

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