8.07 Further applications of trigonometry

Lesson

We have seen a wide variety of trigonometric problems that required us to solve for unknown lengths (distances, heights etc) and angles (including angles of elevation and depression).

Below is a summary of the tools that we may want to use when solving problems involving trigonometry.

Tools for working with right triangles

Pythagorean theorem:  $a^2+b^2=c^2$a2+b2=c2, where c is the hypotenuse

Trigonometric ratios: SOH CAH TOA

$\sin\theta=\frac{\text{Opposite }}{\text{Hypotenuse }}$sinθ=Opposite Hypotenuse = $\frac{O}{H}$OH

$\cos\theta=\frac{\text{Adjacent }}{\text{Hypotenuse }}$cosθ=Adjacent Hypotenuse = $\frac{A}{H}$AH

$\tan\theta=\frac{\text{Opposite }}{\text{Adjacent }}$tanθ=Opposite Adjacent =$\frac{O}{A}$OA

Angle of elevation:  the angle made between the line of sight of the observer and the 'horizontal' when the object is ABOVE the horizontal (observer is looking UP

Angle of depression:  the angle made between the line of sight of the observer and the 'horizontal' when the object is BELOW the horizontal (observer is looking DOWN)

Exact value triangles

We can use these tools in both 2D and 3D. When working in 3D, we will often look at different planes of 2D to see how they fit together.

Worked examples

Question 1

Suppose we have a long rod of marble that we wish to send to a furniture renovator.  We have a box that we need to send it in, and know its dimensions.  What is the longest length of rod that will fit in the box?

Think: The longest rod that can fit in the box with the measurements given is along the diagonal of the box. This is shown on the diagram as$\overline{DE}$DE

To calculate the length of the diagonal, we first need to calculate the length of the diagonal $\overline{BD}$BD using Pythagoras' theorem. This is using the triangle $\triangle ABD$ABD first. Then we can look at $\triangle BDE$BDE to find the desired length.

 $BD^2$BD2 $=$= $AD^2+AB^2$AD2+AB2 State the Pythagorean theorem in terms of the segments given $BD^2$BD2 $=$= $25^2+14^2$252+142 Fill in the given information $BD$BD $=$= $\sqrt{25^2+14^2}$√252+142 Take the square root of both sides $BD$BD $=$= $\sqrt{625+196}$√625+196 Evaluate the squares $BD$BD $=$= $\sqrt{821}$√821 Calculate the sum $BD$BD $=$= $28.65$28.65 cm Round to two decimal places

Now that we have $BD$BD, (the triangle on the base) we can use the Pythagorean theorem again to find length $ED$ED ,(the length of the long diagonal through the box).  This is using triangle $BDE$BDE.

 $DE^2$DE2 $=$= $BE^2+BD^2$BE2+BD2 State the Pythagorean theorem in terms of the segments given $DE^2$DE2 $=$= $16^2+28.65^2$162+28.652 Fill in the given information (for extra accuracy we could use $\sqrt{821}$√821 instead of $28.65$28.65) $DE$DE $=$= $\sqrt{16^2+28.65^2}$√162+28.652 Take the square root of both sides $DE$DE $=$= $\sqrt{256+821}$√256+821 Evaluate the squares $DE$DE $=$= $\sqrt{1077}$√1077 Calculate the sum $DE$DE $=$= $32.82$32.82 cm Round to two decimal place

Reflect: So what we just did, was use the Pythagorean theorem twice, on two separate triangles.

Question 2

Find $x$x in the following diagram,

Think:  In order to find $x$x, we will need to identify some other measurements along the way.  It can be really helpful to make a plan for solving. Below is one possible plan for solving this problem:

1. Calculate $AC$AC using trig ratio sine
2. Calculate $ED$ED, $\frac{AC}{3}$AC3
3. Calculate $x$x, using trig ratio sine

Do:

1. Find length $AC$AC using trig ratio sine

 $\sin23^\circ$sin23° $=$= $\frac{43.6}{AC}$43.6AC​ Fill the given information into the trigonometric ratio $AC$AC $=$= $\frac{43.6}{\sin23^\circ}$43.6sin23°​ Cross multiply to solve for AC $AC$AC $=$= $111.59$111.59 Evaluate on calculator

2. Find length $ED$ED, $\frac{AC}{3}$AC3

 $ED$ED $=$= $\frac{AC}{3}$AC3​ $\overline{ED}$ED is one of three equal segments that make up $\overline{AC}$AC $ED$ED $=$= $\frac{111.59}{3}$111.593​ Fill in known values $ED$ED $=$= $37.2$37.2 Evaluate

3. Find length $x$x, using trig ratio sine

 $\sin35.6^\circ$sin35.6° $=$= $\frac{x}{37.2}$x37.2​ Fill known values into the trigonometric ratio $x$x $=$= $37.2\times\sin35.6^\circ$37.2×sin35.6° Multiply both sides by $37.2$37.2 $x$x $=$= $21.65$21.65

Practice questions

Question 3

A ship is $27$27m away from the bottom of a cliff. A lighthouse is located at the top of the cliff. The ship's distance is $34$34m from the bottom of the lighthouse and $37$37m from the top of the lighthouse.

1. Find the distance from the bottom of the cliff to the top of the lighthouse, $y$y, correct to 2 decimal places.

2. Find the distance from the bottom of the cliff to the bottom of the lighthouse, $x$x, correct to 2 decimal places.

3. Hence find the height of the lighthouse to the nearest tenth of a meter.

Question 4

If $d$d is the distance between the base of the wall and the base of the ladder, find $d$d to two decimal places.

Question 5

Does $\sin\left(90^\circ-\theta\right)=\cos\theta$sin(90°θ)=cosθ?

1. Write down the value of the ratio represented by $\sin\left(90^\circ-\theta\right)$sin(90°θ).

$\sin\left(90^\circ-\theta\right)$sin(90°θ)=$\editable{}$

2. Write down the value of the ratio represented by $\cos\theta$cosθ.

$\cos\theta$cosθ=$\editable{}$

3. Hence, does $\sin\left(90^\circ-\theta\right)=\cos\theta$sin(90°θ)=cosθ?

Yes

A

No

B

Yes

A

No

B

Outcomes

GEO-G.SRT.8

Use sine, cosine, tangent, the Pythagorean Theorem and properties of special right triangles to solve right triangles in applied problems. ★