Solutions to Ratio Problems

We present the solutions to the trigonometric ratio problems.

Solution to Problem 1:

First we need to find the hypotenuse using Pythagora's theorem.

hypotenuse 2 = 8 2 + 6 2 = 100

and hypotenuse = 10

We now use the definitions of the six trigonometric ratios given above to find sin A, cos A, tan A, sec A, csc A and cot A.

sin A = side opposite angle A / hypotenuse = 8 / 10 = 4 / 5

cos (A) = side adjacent to angle A / hypotenuse = 6 / 10 = 3 / 5

tan (A) = side opposite angle A / side adjacent to angle A
= 8 / 6 = 4 / 3

sec (A) = hypotenuse / side adjacent to angle A = 10 / 6
= 5 / 3

csc (A) = hypotenuse / side opposite to angle A
= 10 / 8 = 5 / 4

cot (A) = side adjacent to angle A / side opposite angle A
= 6 / 8 = 3 / 4

Solution to Problem 2:

We are given angle A and the side opposite to it with c the hypotenuse. The sine ratio gives a relationship between the angle, the side opposite to it and the hypotenuse as follows

sin A = opposite / hypotenuse

Angle A and opposite side are known, hence

sin 31 o = 5.12 / c

Solve for c

c = 5.12 / sin 31 o

and use a calculator to obtain

c (approximately) = 9.94

Solution to Problem 3:

If sin x = opposite / hypotenuse = 3 / 7, then we can say that opposite = 3 and hypotenuse = 7 and find the adjacent side using Pythagora's theorem.

hypotenuse 2 = adjacent 2 + opposite 2

7 2 = adjacent 2 + 3 2

adjacent = sqrt (40) = 2 sqrt (10)

We now use trigonometric ratios to find

cos x = adjacent / hypotenuse = 2 sqrt (10) / 7

cot x = adjacent / opposite = 2 sqrt (10) / 3

Solution to Problem 4:

The sine function involves x and the hypotenuse as follows.

sin 30 o = x / 10

Use sin 30 o = 1 / 2 ( see
table of special angles) to find x

x = 5

We now use Pythagora's theorem and write

10 2 = 5 2 + y 2

We now solve for y to obtain

y = SQRT(75) = 5 SQRT (3)

Problem 5: If x is an acute angle and tan x = 5, find the exact value of the trigonometric functions sin x and cos x.

Solution to Problem 5:

If tan x = opposite / adjacent = 5 = 5 / 1, then we can say that opposite = 5 and adjacent = 1 and find the hypotenuse using Pythagora's theorem.

hypotenuse 2 = adjacent 2 + opposite 2

hypotenuse 2 = 1 2 + 5 2

hypotenuse = sqrt (26)

We now use trigonometric ratios to find

sin x = opposite / hypotenuse = 5 / sqrt (26)

cos x = adjacent / hypotenuse = 1 / sqrt (26)



More references on solving problems related to trigonometry and geometry.

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