8.1.3 Solving Right Triangles (continued) Example 3 Solving Right Triangles
In this example, the length of a hypotenuse is calculated by the Pythagorean Theorem, and the measures of two unknown angles in a right triangle are calculated using inverse trigonometric ratios. The lengths of the two legs of the right triangle are given. To find the length of the hypotenuse of the triangle, substitute the lengths of the legs as a and b into a 2 + b 2 = c 2 and solve for c , the length of the hypotenuse. The solution yields a length of approximately 8.07 units. To find m ∠ P , recognize tan P = 6/5.4, and that m ∠ P = tan − 1 (6/5.4) ≈ 48 ° . Since ∠ R is the supplement of ∠ P in the right triangle, its measure is approximately 90 ° − 48 ° = 42 ° .
Example 4 Solving a Right Triangle in the Coordinate Plane
The side lengths and angle measures of a triangle in the coordinate plane are determined in this example. The coordinates of the vertices are given. The lengths of the two legs of the triangle can be obtained by subtracting the x coordinates of the endpoints for the horizontal leg and subtracting the y coordinates of the endpoints for the vertical leg. The length of the hypotenuse is calculated using the Distance Formula. The lengths are KL = 5, JK = 12, and JL = 13; this is a Pythagorean triple. Since JK and KL are perpendicular (they are each parallel to an axis of the coordinate plane), m ∠ K = 90 ° . To obtain the measure of ∠ L , take the inverse tangent of the ratio of the opposite leg over the adjacent leg: m ∠ L = tan − 1 (12/5) ≈ 67 ° . The final angle, ∠ J , is complementary to ∠ L because they are the acute angles in a right triangle: m ∠ J = 90 ° − 67 ° ≈ 23 ° .
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