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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