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Trigonometry of Right Triangles Right Triangles Tamara Kucherenko Tamara Kucherenko Right Triangles Trigonometry of Right Triangles Trigonometric Ratios We define sine, cosine, tangent, cosecant, secant, cotangent as hypotenuse opposite θ adjacent The Trigonometric Ratios sin θ = opposite hypotenuse cos θ = adjacent hypotenuse tan θ = opposite adjacent csc θ = hypotenuse opposite sec θ = hypotenuse adjacent cot θ = adjacent opposite The trigonometric ratios depend only on the angle θ and not on the size of the triangle. The reason is that any two right triangles with angle θ are similar and the ratios of their corresponding sides are the same. Tamara Kucherenko Right Triangles Trigonometry of Right Triangles Trigonometric Ratios Example 1. Find the six trigonometric ratios of the angle θ. √ 13 2 θ 3 Tamara Kucherenko Right Triangles Trigonometry of Right Triangles Trigonometric Ratios Example 1. Find the six trigonometric ratios of the angle θ. 2 3 2 Solution. sin θ = √ cos θ = √ tan θ = 3 13 13 √ √ 13 13 3 csc θ = sec θ = cot θ = 2 3 2 Tamara Kucherenko Right Triangles √ 13 2 θ 3 Trigonometry of Right Triangles Trigonometric Ratios Example 1. Find the six trigonometric ratios of the angle θ. 2 3 2 Solution. sin θ = √ cos θ = √ tan θ = 3 13 13 √ √ 13 13 3 csc θ = sec θ = cot θ = 2 3 2 Example 2. If sin θ = 47 , sketch a right triangle with acute angle θ. Tamara Kucherenko Right Triangles √ 13 2 θ 3 Trigonometry of Right Triangles Trigonometric Ratios Example 1. Find the six trigonometric ratios of the angle θ. 2 3 2 Solution. sin θ = √ cos θ = √ tan θ = 3 13 13 √ √ 13 13 3 csc θ = sec θ = cot θ = 2 3 2 √ 2 θ 3 Example 2. If sin θ = 47 , sketch a right triangle with acute angle θ. Solution. Since sin θ is dened as the ratio of the opposite side to the hypotenuse, we sketch a triangle with hypotenuse of length 7 and a side of length 4 opposite to θ. Tamara Kucherenko Right Triangles 13 7 4 θ Trigonometry of Right Triangles Trigonometric Ratios Example 1. Find the six trigonometric ratios of the angle θ. 2 3 2 Solution. sin θ = √ cos θ = √ tan θ = 3 13 13 √ √ 13 13 3 csc θ = sec θ = cot θ = 2 3 2 √ 2 θ 3 Example 2. If sin θ = 47 , sketch a right triangle with acute angle θ. Solution. Since sin θ is dened as the ratio of the opposite side to the hypotenuse, we sketch a triangle with hypotenuse of length 7 and a side of length 4 opposite to θ. To find the adjacent side we use the Pythagorean Theorem: √ (adjacent)2 = 72 − 42 = 33, so adjacent = 33. Tamara Kucherenko Right Triangles 13 7 4 θ √ 33 Trigonometry of Right Triangles Special Triangles We can use the special triangles below to calculate the trigonometric ratios for angles with measures 30◦ , 45◦ , and 60◦ . √ 2 1 2 30◦ √ 45◦ 1 Tamara Kucherenko 60◦ 1 Right Triangles 3 1 Trigonometry of Right Triangles Special Triangles We can use the special triangles below to calculate the trigonometric ratios for angles with measures 30◦ , 45◦ , and 60◦ . √ 2 30◦ 2 1 √ 45◦ 60◦ 1 1 3 1 θ in degrees θ in radians sin θ cos θ tan θ csc θ sec θ 30◦ π 6 π 4 π 3 1 √2 2 √2 3 2 3 √2 3 2 1 2 3 3 2 √ 2 √ 45◦ 60◦ √ √ 1 √ 3 √ 2 3 3 √ 2 3 3 2 2 cot θ √ 3 1 √ 3 3 You do not need to, and should not, memorize this table. Rather, memorize the two red triangles and be able to quickly calculate all trig functions of 30, 45, or 60 degrees. Tamara Kucherenko Right Triangles Trigonometry of Right Triangles Applications of Trigonometry of Right Triangles To solve a triangle means to determine all of its parts from the information known about the triangle, that is, to determine the lengths of the three sides and the measures of the three angles. Example. Solve the right triangle B 5 A π 3 C Tamara Kucherenko Right Triangles Trigonometry of Right Triangles Applications of Trigonometry of Right Triangles To solve a triangle means to determine all of its parts from the information known about the triangle, that is, to determine the lengths of the three sides and the measures of the three angles. Example. Solve the right triangle B 5 A π 3 π π π − = . 2 3 6 To find AC, we look for an equation that relates AC to the lengths and angles we already know. In AC this case, we have sin π3 = AB = AC 5 . Thus, Solution. First, ∠A = √ π 5 3 AC = 5 sin = . 3 2 C Similarly, BC = 5 cos Tamara Kucherenko Right Triangles π 5 = . 3 2