Given a right triangle with one leg of length and hypotenuse of length , what is the approximate length of the other leg?
Table of contents
- 0. Review of College Algebra4h 43m
- 1. Measuring Angles40m
- 2. Trigonometric Functions on Right Triangles2h 5m
- 3. Unit Circle1h 19m
- 4. Graphing Trigonometric Functions1h 19m
- 5. Inverse Trigonometric Functions and Basic Trigonometric Equations1h 41m
- 6. Trigonometric Identities and More Equations2h 34m
- 7. Non-Right Triangles1h 38m
- 8. Vectors2h 25m
- 9. Polar Equations2h 5m
- 10. Parametric Equations1h 6m
- 11. Graphing Complex Numbers1h 7m
2. Trigonometric Functions on Right Triangles
Solving Right Triangles
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Join thousands of students who trust us to help them ace their exams!Watch the first videoMultiple Choice
Point P is the center of the circle in the figure above. If triangle is a right triangle with right angle at , and , , what is the value of if ?
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Verified step by step guidance1
Identify the given elements: triangle \( A P B \) is a right triangle with the right angle at \( A \), and the lengths \( A P = 5 \) and \( P B = 13 \). We need to find \( A B = x \).
Recall the Pythagorean theorem for a right triangle: the square of the hypotenuse equals the sum of the squares of the other two sides. Since the right angle is at \( A \), the side opposite \( A \) (which is \( P B \)) is the hypotenuse.
Set up the Pythagorean theorem equation: \( (A P)^2 + (A B)^2 = (P B)^2 \). Substitute the known values: \( 5^2 + x^2 = 13^2 \).
Simplify the equation: \( 25 + x^2 = 169 \).
Isolate \( x^2 \) and solve for \( x \): \( x^2 = 169 - 25 \), then \( x = \sqrt{144} \).
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