Given a right triangle with sides $a$, $b$, and hypotenuse $c$, which equation can be used to solve for $c$ using the Law of Cosines?

• A. $c^2=a^2+b^2+2abcos90°$
• B. $c^2=a^2-b^2$
• C. $c^2=a^2+b^2$
• D. $c^2=a^2+b^2-2abcos90°$

Answer: (D)

Verified step by step guidance

1. Recall the Law of Cosines formula for any triangle with sides a, b, and c, and the angle \( \theta \) opposite side c: \[ c^{2} = a^{2} + b^{2} - 2ab \cos(\theta) \] This formula generalizes the Pythagorean theorem and accounts for any angle \( \theta \).
2. Identify that in a right triangle, the angle \( \theta \) opposite the hypotenuse \( c \) is \( 90^\circ \). Substitute \( \theta = 90^\circ \) into the Law of Cosines formula: \[ c^{2} = a^{2} + b^{2} - 2ab \cos(90^\circ) \]
3. Recall the value of \( \cos(90^\circ) \), which is 0. This simplifies the equation because the term \( -2ab \cos(90^\circ) \) becomes zero: \[ c^{2} = a^{2} + b^{2} - 2ab \times 0 \]
4. Simplify the equation by removing the zero term: \[ c^{2} = a^{2} + b^{2} \] This is the Pythagorean theorem, which is a special case of the Law of Cosines for right triangles.
5. Therefore, the Law of Cosines equation used to solve for \( c \) in a right triangle is: \[ c^{2} = a^{2} + b^{2} - 2ab \cos(90^\circ) \] which simplifies to the Pythagorean theorem due to \( \cos(90^\circ) = 0 \).