6.09 Problem solving with trigonometry

To find missing angles in right-angled triangles using trigonometry, we require at least two known sides and the trigonometric ratios.

Remember that the trigonometric ratios for a right-angled triangle are:\sin \theta =\dfrac{\text{Opposite }}{\text{Hypotenuse }} \quad \quad \cos \theta =\dfrac{\text{Adjacent }}{\text{Hypotenuse }} \quad \quad \tan \theta =\dfrac{\text{Opposite }}{\text{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(\dfrac{\text{Opposite }}{\text{Hypotenuse }}\right) \quad \quad \theta =\cos^{-1}\left(\dfrac{\text{Adjacent }}{\text{Hypotenuse }}\right) \quad \quad \theta =\tan^{-1}\left(\dfrac{\text{Opposite }}{\text{Adjacent }}\right)

Any of these relationships can be used to find \theta depending on which side lengths of the triangle are known.

While we may represent the inverse trigonometric functions using an index of -1, they are not reciprocals of the original functions.

For example: \sin^{-1} \theta \neq \dfrac{1}{\sin \theta}

To find missing sides is right-angled triangles using trigonometry, we require at least one angle (other than the right angle) and at least one other side length.

We can find a missing side length by expressing it in a trigonometric ratio as the only unknown value, and then solve the equation to find that value.

Depending on the given angle, given side and missing side, we will need to use one of the three trigonometric ratios so that all relevant information is in the equation.

Examples

Example 1

AB is a tangent to a circle with centre O. OB is 20 cm long and cuts the circle at C.

Find the length of BC to two decimal places.

Worked Solution

Create a strategy

Use the cosine ratio.

Apply the idea

Notice that we can find the length of BC by subtracting the length of OC from OB. Also, we can see that OC has the same length as OA as they are both radii. So we need to calculate the OA first.

With respect to the given angle of 52\degree, the adjacent side is OA, and the hypotenuse is OB=20, so we can use the cosine ratio.

\displaystyle \cos \theta	\displaystyle =	\displaystyle \frac{\text{Adjacent }}{\text{Hypotenuse}}	Use the cosine ratio
\displaystyle \cos 52\degree	\displaystyle =	\displaystyle \frac{OA}{20}	Substitute the values
\displaystyle OA	\displaystyle =	\displaystyle 20\cos 52\degree	Multiply both sides by 20

\displaystyle BC	\displaystyle =	\displaystyle OB-OC	Subtract OC from OB
	\displaystyle =	\displaystyle 20-20\cos 52\degree	Substitute the values
	\displaystyle \approx	\displaystyle 7.69 \text{ cm}	Evaluate using a calculator

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In some cases, a single application of trigonometry on a single right-angled triangle will not be enough to find the values we are looking for.

In these cases, we can use trigonometry once to find a new value, and then use trigonometry again with our new value to find another new value. We can repeat this process as many times as is necessary to find the value that we are looking for.

Examples

Example 2

Find the size of angle z. Give the answer in degrees, minutes and seconds, rounding to the nearest second.

Worked Solution

Create a strategy

Use the inverse tangent function.

Apply the idea

With respect to the angle of x, the length of the opposite side is 20, and the length of the adjacent side is 50, so we can use the inverse tangent function.

\displaystyle x	\displaystyle =	\displaystyle \tan^{-1}\left(\dfrac{\text{Opposite }}{\text{Adjacent }}\right)	Use the inverse \tan function
\displaystyle x	\displaystyle =	\displaystyle \tan^{-1}\left(\dfrac{10}{35}\right)	Substitute the values
	\displaystyle \approx	\displaystyle 15.9454\degree	Evaluate

We do the same process to find the angle of y, which has the opposite side length of 50, and the adjacent side length of 25.

\displaystyle y	\displaystyle =	\displaystyle \tan^{-1}\left(\dfrac{35}{15}\right)	Substitute the values
	\displaystyle \approx	\displaystyle 66.8014\degree	Evaluate

Since the angle sum of a triangle is 180\degree, we know x+y+z+90= 180.

\displaystyle z	\displaystyle =	\displaystyle 180-90-x-y	Subtract 90,\,x, and y from both sides
	\displaystyle =	\displaystyle 180-90-15.9454-66.8014	Substitute x and y
	\displaystyle =	\displaystyle 7.2532\degree	Evaluate using a calculator
	\displaystyle =	\displaystyle 7\degree + 0.2532 \times 60 \text{ minutes}	Convert the decimal to minutes
	\displaystyle =	\displaystyle 7\degree 15.192\rq	Evaluate the multiplication
	\displaystyle =	\displaystyle 7\degree 15\rq + 0.192 \times 60 \text{ seconds}	Convert the decimal to seconds
	\displaystyle \approx	\displaystyle 7 \degree 15 \rq 12\rq\rq	Evaluate the multiplication

In other cases, multiple applications of trigonometry need to be used at the same time in order to solve for a common factor.

These cases arise when multiple trigonometric ratios involve the same missing value. If some relationship between these ratios is known then the missing value can be solved for.

Examples

Example 3

Find the length of AB to two decimal places.

Worked Solution

Create a strategy

Use the tangent ratio.

Apply the idea

With respect to the given angle of 58\degree, the length of the opposite side is CD=14, and the adjacent side is AD. With respect to the given angle of 29\degree, the length of the opposite side is CD=14, and the adjacent side is DB. So we can use the tangent ratio to calculate the length of AD and DB.

We can use the rearranged trigonometric ratio of tangent to find both distances and substitue the values: \text{Adjacent}=\dfrac{\text{Opposite}}{\tan \theta}

AD=\dfrac{14}{\tan 58\degree} \quad \quad DB=\dfrac{14}{\tan 29\degree}

So we have:

\displaystyle AB	\displaystyle =	\displaystyle \dfrac{14}{\tan 58\degree}+\dfrac{14}{\tan 29\degree}	Add the two adjacent sides
	\displaystyle =	\displaystyle 25.256669 + 8.748171	Evaluate using a calculator
	\displaystyle \approx	\displaystyle 34.00	Evaluate the addition

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