Question

which equation can be used to solve for x in the triangle below?
triangle with vertices q, p, r; pq = 4.0 cm, pr = 4.7 cm, angle at p is 41°, qr = x
options:

• $x^2 = 4.7^2 + 4.0^2 + 2(4.7)(4.0)cos41°$
• $x = 4.7 + 4.0 + 2(4.7)(4.0)cos41°$
• $x^2 = 4.7^2 + 4.0^2 - 2(4.7)(4.0)cos41°$
• $x = 4.7 + 4.0 - 2(4.7)(4.0)cos41°$

Explanation:

Step1: Recall the Law of Cosines

The Law of Cosines states that for a triangle with sides \(a\), \(b\), and \(c\), and the angle \(\theta\) opposite side \(c\), the formula is \(c^{2}=a^{2}+b^{2}-2ab\cos\theta\).

Step2: Identify the sides and angle in the triangle

In triangle \(PQR\), we have two sides: \(a = 4.7\) cm, \(b = 4.0\) cm, and the included angle \(\theta=41^{\circ}\) between them. The side we want to find is \(x\), which is opposite the angle \(41^{\circ}\) (or we can consider \(x\) as \(c\) in the Law of Cosines formula).

Step3: Apply the Law of Cosines to the triangle

Substituting \(a = 4.7\), \(b = 4.0\), \(\theta = 41^{\circ}\), and \(c=x\) into the Law of Cosines formula \(c^{2}=a^{2}+b^{2}-2ab\cos\theta\), we get \(x^{2}=4.7^{2}+4.0^{2}-2(4.7)(4.0)\cos41^{\circ}\).

Answer:

\(x^{2}=4.7^{2}+4.0^{2}-2(4.7)(4.0)\cos41^{\circ}\) (the third option)