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6.02 Trigonometric ratios

Lesson

The trigonometric ratios

The trigonomteric ratios describe the relationship between a given angle and two sides of a right-angled triangle. These rely on the trigonometric functions $\sin$sin, $\cos$cos and $\tan$tan.

The trigonometric ratios are:

  • $\sin\theta=\frac{\text{Opposite }}{\text{Hypotenuse }}$sinθ=Opposite Hypotenuse
  • $\cos\theta=\frac{\text{Adjacent }}{\text{Hypotenuse }}$cosθ=Adjacent Hypotenuse
  • $\tan\theta=\frac{\text{Opposite }}{\text{Adjacent }}$tanθ=Opposite Adjacent

To isolate the angle in each of these relationships, we can apply the inverse trigonometric function to each side of the equation. This will give us:

  • $\theta=\sin^{-1}\left(\frac{\text{Opposite }}{\text{Hypotenuse }}\right)$θ=sin1(Opposite Hypotenuse )
  • $\theta=\cos^{-1}\left(\frac{\text{Adjacent }}{\text{Hypotenuse }}\right)$θ=cos1(Adjacent Hypotenuse )
  • $\theta=\tan^{-1}\left(\frac{\text{Opposite }}{\text{Adjacent }}\right)$θ=tan1(Opposite Adjacent )

If we are given any two values in a right-angled triangle, either angles or side lengths, we can use these ratios to find any other angles or side lengths in the triangle.

 

Finding a side

Based on where the angle is in the triangle and which pair of sides we are working with, we can choose one of the trigonometric ratios to describe the relationship between those values.

We can then rearrange that ratio to make our unknown value the subject of an equation and then evaluate to find its value.

Worked example

Find the value of $x$x.

Think: With respect to the given angle, the side of length $5$5 is adjacent and the side of length $x$x is opposite. This means that we should choose the trigonometric ratio $\tan\theta=\frac{\text{Opposite }}{\text{Adjacent }}$tanθ=Opposite Adjacent to relate our given values.

Do: Substituting our given values into the trigonometric ratio gives us:

$\tan38^\circ=\frac{x}{5}$tan38°=x5

We can multiply both sides of the equation by $5$5 to make $x$x the subject and then evaluate to find its value (rounded to two decimal places):

$x=5\tan38^\circ$x=5tan38°$=$=$3.91$3.91

Reflect: After identifying which sides we were working with, we chose the trigonometric ratio that matched those sides. We then solved the equation to find our unknown side length.

 

Finding an angle

Based on where the angle is in the triangle and which pair of sides we are given, we can choose one of the trigonometric ratios to describe the relationship between those values.

We can then rearrange that ratio (or choose the corresponding inverse ratio) to make our unknown angle the subject of an equation and then solve for it.

Worked example

Find the value of $x$x.

Think: With respect to the angle $x$x, the side of length $4$4 is opposite and the side of length $5$5 is adjacent. This means that we can use the inverse trigonometric ratio $\theta=\tan^{-1}\left(\frac{\text{Opposite }}{\text{Adjacent }}\right)$θ=tan1(Opposite Adjacent ).

Do: Substituting our values into the inverse trigonometric ratio gives us:

$x=\tan^{-1}\frac{4}{5}$x=tan145

Evaluating $x$x (and rounding to two decimal places) gives us:

$x=38.66$x=38.66

Reflect: After identifying which sides we were given, we chose the inverse trigonometric ratio that matched those sides. We then solved the equation to find our unknown angle size.

 

Practice questions

Question 1

Consider the angle $\theta$θ.

What is the value of the ratio $\frac{\text{Opposite }}{\text{Hypotenuse }}$Opposite Hypotenuse ?

Question 2

Consider the following triangle.

  1. Find the value of $x$x.

  2. Hence find the value of $\sin\theta$sinθ.

  3. Hence find the value of $\cos\theta$cosθ.

Question 3

Given $\tan\theta=\frac{3}{4}$tanθ=34, find the value of $b$b.

Question 4

Consider the following figure.

  1. Find the value of $x$x.

    Round your answer to two decimal places.

  2. Next, find the value of $y$y.

    Round your answer to two decimal places.

  3. Hence or otherwise, find the length of the base of the triangle.

    Round your answer to two decimal places.

Outcomes

VCMMG346

Solve right-angled triangle problems including those involving direction and angles of elevation and depression.

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