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Exercises 20.5 Applications of the Pythagorean identities 1 A horizontal force P holds a simple pendulum inclined at an angle \theta to the vertical. Given that $T \cos \theta = mg$ and $T \sin \theta = P$, show that $(mg)^2 + P^2 = T^2$. Hence evaluate T, the tension in the string, when $m = 2.46$ kg and $P = 17.6$ N ($g = 9.80$ m/s$^2$).

          Exercises 20.5
Applications of the Pythagorean identities
1 A horizontal force P holds a simple pendulum inclined at an
angle \theta to the vertical. Given that $T \cos \theta = mg$ and $T \sin \theta = P$,
show that $(mg)^2 + P^2 = T^2$. Hence evaluate T, the tension in
the string, when $m = 2.46$ kg and $P = 17.6$ N ($g = 9.80$ m/s$^2$).

Exercises 20.5
Applications of the Pythagorean identities
1 A horizontal force P holds a simple pendulum inclined at an
angle θto the vertical. Given that T cosθ = mg and T sinθ = P,
show that (mg)^2 + P^2 = T^2. Hence evaluate T, the tension in
the string, when m = 2.46 kg and P = 17.6 N (g = 9.80 m/s^2).

Added by Bobby H.

Introductory and Intermediate Algebra for College Students 4th

Robert Blitzer 4th Edition

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Solved by Expert Timothy James

01/02/2023

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Step 1: Start with the given equations T cos θ = mg and T sin θ = P. Show more…

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Texts: Exercises 20.5 Applications of the Pythagorean identities 1. A horizontal force P holds a simple pendulum inclined at an angle to the vertical. Given that cos θ = mg and T sin θ = P, show that (mg)² + P² = T². Hence, evaluate T, the tension in the string, when m = 2.46 kg and P = 17.6 N (g = 9.80 m/s²). Happy with getting the result, cannot understand how to prove it. Exercises 20.5 Applications of the Pythagorean identities A horizontal force P holds a simple pendulum inclined at an angle to the vertical. Given that T cos θ = mg and T sin θ = P, show that (mg)² + P² = T². Hence, evaluate T, the tension in the string when m = 2.46 kg and P = 17.6 N (g = 9.80 m/s²).