5.02 Applications using right triangle trigonometry
Applications with angles of elevation and depression
In real-life contexts we use special words to describe particular angles.
Angle of elevation
An angle of elevation is the angle created when an observer is looking at a object which is above the horizontal. The angle between the horizontal and the observer's line of sight is called angle of elevation. We may also see this referred to as the angle of inclination.
Angle of depression
An angle of depression is the angle created when an observer is looking at an object which is below the horizontal. The angle between the horizontal and the observer's line of sight is called angle of depression.
Using the angle of elevation or depression, we can create right triangles. If we know the height above or below we can find the distance an object is from an observer, or vice-versa.
Angles of elevation and depression are common in applications of trigonometry. Consider how we could find the height to the peak of a mountain without needing to climb it, or how we could find the height of an airplane in the sky. Trigonometry can help with a lot of these. In fact there are many examples of professions that use trigonometry:
- Surveyors measure an angle and a length, then use right triangle trigonometry to get another length that cannot be measured directly, such as a distance across a lake.
- Astronomers measure the lengths of shadows of mountains on the moon and knowing the sun angle can compute the heights of the mountains. They also use it when finding the distance between celestial bodies
- Architects use trigonometry to calculate structural load, roof slopes, ground surfaces and many other aspects, including sun shading and light angles
- Ship Captains use trigonometry in navigation to find the distance of the shore from a point in the sea.
- Oceanographers use trigonometry when calculating the height of tides in oceans
Worked example
Question 1
Jasper and Jasmine were playing in the park and wondered how tall the tallest tree would be. Jasper tried climbing the tree, trying to use his tape measure as he climbed. This took a few hours, and was quite tricky in places. Jasmine remembered some stuff about trigonometry from school and measured the distance from the tree and the angle of elevation. These measurements were much easier to get and didn't involve any climbing!
Think: The distance from the base of the tree to a point of observation was $4.2$4.2 m. The angle is above the horizon so, it is an angle of elevation from this point to the top of the tree and is $38^\circ$38°.
Do:
| $\tan\theta$tanθ | $=$= | $\frac{O}{A}$OA |
| $\tan38^\circ$tan38° | $=$= | $\frac{\text{height of tree }}{4.2}$height of tree 4.2 |
| $\text{height of tree }$height of tree | $=$= | $4.2\times\tan38^\circ$4.2×tan38° |
| $\text{height of tree }$height of tree | $=$= | $3.28$3.28 |
Reflect: So the height of this tree is $3.28$3.28 m.
Practice questions
Question 2
At a certain time of the day a light post, $6$6 m tall, has a shadow of $5.8$5.8 m. If the angle of elevation of the sun at that time is $\theta$θ°, find $\theta$θ to two decimal places.
Question 3
A fighter jet, flying at an altitude of $2000$2000 m is approaching a target. At a particular time the pilot measures the angle of depression to the target to be $13^\circ$13°. After a minute, the pilot measures the angle of depression again and finds it to be $16^\circ$16°.
The path of a fighter jet to the airport is depicted as $AC$AC. Point $B$B lies on $AC$AC. A vertical line segment connects point $C$C to a point labeled "Airport" below. Three line segments, including the vertical line segment, connect points $A$A, $B$B, and $C$C to Airport. The line segments drawn formed triangles. The first triangle has vertices $A$A, $C$C, and Airport. The second triangle has vertices has vertices $B$B, $C$C, and Airport. The third triangle has vertices $A$A, $B$B, and Airport. $AC$AC is labeled to measure 2000 m. Point $A$A represents the initial location of the fighter jet, while point $B$B represents the location of the fighter jet after a minute. The angle between $AC$AC and the line connecting point $A$A to "Airport" is marked by a blue marking and measures $13^\circ$13°. The angle between $BC$BC and the line connecting point $B$B to "Airport" is marked by a purple marking and measures $16^\circ$16°.
Find the distance $AC$AC.
Round your answer to the nearest meter.
Find the distance $BC$BC.
Round your answer to the nearest meter.
Now find the distance covered by the jet in one minute.
Round your answer to the nearest meter.