1. Lesson 3: Solving Problems Involving Right Triangles
Trigonometry
Lesson 3: Solving Problems Involving Right Triangles

2. Todays Objectives
Students will be able to develop and apply the primary trigonometric ratios (sine, cosine, tangent) to solve problems that involve right triangles, including:
Solve right triangles, with or without technology

3. Solving Problems Involving Right Triangles
There are many applications of trigonometry and the Pythagorean theorem, especially in building construction, engineering, and navigation.
Since diagrams are often not provided, the key to solving many problems is to sketch and accurately label a diagram

4. Example
A 12-foot ladder is leaning against a vertical wall. If it reaches 9.5 ft. up the wall, what angle does it make with the level ground, to the nearest degree?
Solution:
First, draw and label a sketch of a representative
diagram that includes a right triangle.
The required angle is labeled θ. The measure of
the angle θ can be determined from the sine ratio.
sin𝜃 = 𝑜𝑝𝑝 ℎ𝑦𝑝 =
𝜃 = sin −
𝜃 = 52º
The ladder makes an angle of approximately 52º with the ground.
12 ft
9.5 ft 𝜃

5. Angles of Elevation and Depression
The ladder problem is an example of an angle of elevation
An angle of elevation is measured between two rays that have a common starting point called the vertex of the angle. One ray is horizontal, while the other is above the horizontal.
Vertex
Vertex
Angle of elevation
Horizontal Ray
Also called Angles of Inclination
Horizontal Ray

6. Angles of Elevation and Depression
An angle of depression is similar to an angle of elevation, except that the second ray is below the horizontal ray.

7. Compass Directions and Headings
Compass directions may be given as north (N), south (S), east (E), and west (W); or, halfway between these, such as northwest (NW).
Directions can also be given as the number of degrees east or west, north or south. This is called a heading.
For headings, north or south is listed first.
For example, a heading of N40ºE means 40º east of due north, as shown in this illustration. N
N40°E
40° W E S

8. Bearings
A bearing is a three-digit number of degrees between 000º and 360º that indicates a direction measured clockwise from north. For example, a bearing of 127º is in a direction 127º measured clockwise from due north, as shown below.
127° N
Bearing of 127°

9. Example
From a vertical lighthouse 200 ft. above the surface of the water, a person sees a sailboat at an angle of depression of 27º. How far is the sailboat from the water level of the lighthouse, to the nearest foot?
Solution: First, label the diagram
The angle of depression from the lighthouse to the sailboat is equal to the angle of elevation from the sailboat to the lighthouse
200 ft
27° x
27°

10. Example Determine x by applying the tangent ratio. tan 27 º = 𝑜𝑝𝑝 𝑎𝑑𝑗
The sailboat is approximately 393 ft. from the water level of the lighthouse.

11. Example (You do)
A xiaomao is watching a bird in a tree. From the position of the cat, the bird is at an angle of elevation of 40º. If the cat is 7.1 m from the base of the tree, how high up in the tree is the bird, to the nearest tenth of a meter?
Solution: draw and label a diagram
40°
7.1 m x

12. Solution Determine x by applying the tangent ratio.
The height of the bird is approximately 6 meters.

13. Example
A ship travels from a port at 30 km/h on a bearing of 137º. How far east of the port is the ship after three hours, to the nearest tenth of a kilometer?
Solution:
After three hours the ship has gone 3 x 30 km = 90 km. This is shown in the given labeled diagram.
90 km
137° x
Note that the acute angle between the vertical line and the direction of travel is 180°- 137°= 43°
43°
𝑠𝑖𝑛43= 𝑜𝑝𝑝 ℎ𝑦𝑝 = 𝑥 90
90𝑠𝑖𝑛43=𝑥=61.4 𝑘𝑚

14. Example (You do)
An airplane flies for 2.5 hours at 240 km/h on a heading of S25ºW. How far south has the plane flown during this time, to the nearest km?
Solution:
First, draw and label a sketch diagram of a representative diagram that includes a right triangle. Let x be the unknown distance south. N
After 2.5 hours the airplane has flown 2.5 x 240 km = 600 km
Determine x by applying the cosine ratio
𝑐𝑜𝑠25= 𝑎𝑑𝑗 ℎ𝑦𝑝 = 𝑥 600
𝑥=600𝑐𝑜𝑠25=544 𝑘𝑚
The airplane has flown approx. 544 km south
25° x
600
S25°W