Trigonometry and triangles
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Trigonometry deals with relationships between the sides and angles of triangles. It uses trigonometric functions like sine, cosine, and tangent. Sine relates the opposite side to the hypotenuse, cosine relates the adjacent side to the hypotenuse, and tangent relates the opposite side to the adjacent side. Trigonometry can be used to find unknown side lengths or angles in triangles, including solving "story problems" where trigonometric applications are needed to determine real-world quantities like shadow lengths. Key aspects of using trigonometry include identifying what quantities are given and needed to be found, setting up the appropriate trigonometric function, and drawing a right triangle to visualize the problem and solve it.
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This document discusses right triangle trigonometry. It defines the six trigonometric functions - sine, cosine, tangent, cotangent, secant, and cosecant - which are ratios of the sides of a right triangle. The three sides are the hypotenuse, which is opposite the right angle, and the opposite and adjacent sides in relation to the acute angle. The document also provides examples of using trigonometric functions to find missing angles and sides of right triangles, including using special right triangle theorems for 45-45-90 and 30-60-90 triangles.
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- Using trigonometric functions to solve practical problems involving angles of elevation and depression
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The document provides a review of trigonometric ratios such as sine, cosine, and tangent that are used to solve problems involving right triangles, includes examples of applying these ratios to solve practice problems involving missing side lengths and angle measures, and reviews special right triangles and the SOH CAH TOA method for remembering the trigonometric functions.
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- Rectangles have opposite parallel sides of equal length and right angles. Squares are rectangles with all sides of equal length.
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- Squares are rectangles with all sides of equal length. The perimeter and area formulas are the same but use only one side length.
- Parallelograms have opposite parallel sides of equal length and opposite interior angles that sum to 180 degrees. The
Right Triangle Trigonometry
This document summarizes key concepts from a chapter on trigonometric functions. It defines the six trig functions (sine, cosine, tangent, cosecant, secant, cotangent) in terms of right triangles and provides examples of how to use trig functions to find unknown side lengths. It also introduces special right triangles used to find exact values of trig functions, such as the 45-45-90 triangle used to find sine and cosine of 45 degrees.
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Trigonometry and triangles
- 1. Trigonometry and Triangles Michael Schmidt
- 2. Trigonometry Sine Cosine Tangent Length of triangle legs Angle of triangle corners Area of triangles What we are doing
- 3. Branch in Mathematics Uses trig functions Triangles Mostly right triangles Uses relationships to find unknowns Trigonometry
- 4. θ (Theta) Adjacent leg (A) Opposite leg (O) Hypotenuse (H) Key Terms H O θ A
- 5. SOH: sin θ =𝑂𝑝𝑝𝑜𝑠𝑖𝑡𝑒𝐻𝑦𝑝𝑜𝑡𝑒𝑛𝑢𝑠𝑒 CAH: cos θ =𝐴𝑑𝑗𝑎𝑐𝑒𝑛𝑡𝐻𝑦𝑝𝑜𝑡𝑒𝑛𝑢𝑠𝑒 TOA: tan θ =𝑂𝑝𝑝𝑜𝑠𝑖𝑡𝑒𝐴𝑑𝑗𝑎𝑐𝑒𝑛𝑡 SOH CAH TOA
- 6. sin 37° = cos37° = tan 37° = SOH CAH TOA continued 21𝑓𝑡35𝑓𝑡 28𝑓𝑡35𝑓𝑡 21𝑓𝑡28𝑓𝑡 35ft 21ft 37° 28ft
- 7. What is given? Which trig function? Finding the side length 12ft sin 30° = 𝑋12𝑓𝑡 X sin 30°1 = 𝑋12𝑓𝑡 X= 6ft 30°
- 8. Using trig, find unknown tan 20° = 𝑋6𝑚 1 X X= 2.18m 20° 6m X cos 45° = 8𝑖𝑛𝑥 1 45° X= 11.31in 8in
- 9. What is known? tan θ = 34 Use 𝑡𝑎𝑛−1 𝑡𝑎𝑛−1( 34 ) = θ θ = 36.86° Using trig to find θ 3’ θ 4’
- 10. sin θ = 10𝑚15𝑚 𝑠𝑖𝑛−1(10𝑚15𝑚 ) = θ θ = 41.81° Solve for θ θ 15m 10m cos θ = 7𝑐𝑚9𝑐𝑚 𝑐𝑜𝑠−1(7𝑐𝑚9𝑐𝑚 ) = θ θ = 38.94° 9cm θ 7cm
- 11. Area of triangle A = 𝐵𝐻2 A = 15∗102 = 75𝑚2 Finding the Area 10m 15m
- 12. What is given? What is needed? How is it found? tan60 = 𝐵10 A =17.32𝑐𝑚∗10𝑐𝑚2 Finding area with trig 60° 10cm B =17.32cm =86.6𝑐𝑚2
- 13. Given Needed sin50 = 𝐻11 A= 13∗8.432 Non right triangles H= 8.43in 11in A=54.8𝑖𝑛2 50° 13in
- 14. cos 60 = 𝐻22 Pythagorean theorem for the base A= 19.05∗112 Find area of triangle H=11in 60° 22in A=104.78𝑖𝑛2
- 15. Given Needed B= X+Y tan 45 = 𝑥16 tan 30 = 𝑦16 B=25.24cm Find area of triangle continued Height = 16cm 16cm 45° 30° X Y X=16cm Y=9.24cm A=16∗25.242 =201.92𝑐𝑚2
- 16. A 6ft tall man is standing in front of a light. The light is casting a shadow. If the angle of depression at the man’s head is 60° how long is the shadow? Story Problems tan 60 = 𝐿6 L=10.39ft 60° 6ft L
- 17. Story problems There is a window 33ft up a building and the only ladder is 40ft long. For safety reasons the ladder is leaned against the building at 52°. Will the ladder reach the window? sin 52 = 𝐻40 H=31.52ft 40ft No, the ladder will not reach the window. 52°
- 18. SOH CAH TOA is key Find the Given and Needed Make own right triangle Draw a picture Wrap up
Editor's Notes
- There is a couple ways to do the first problem. Then can use Pythagorean theorem.