Question

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find ab.
round to the nearest tenth.
triangle diagram: vertices b, c, a; bc = 16, ac = 5, angle at c is 61°
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solving for a side with the law of cosines 4:38
solving for an angle with the law of cosines 6:41

Explanation:

Step1: Identify the Law of Cosines

We use the Law of Cosines for triangle \(ABC\) where we know two sides and the included angle. The Law of Cosines formula is \(c^{2}=a^{2}+b^{2}-2ab\cos(C)\), where \(a\), \(b\) are the sides and \(C\) is the included angle. Here, we want to find \(AB\) (let's call it \(c\)), \(a = 16\), \(b = 5\), and \(C=61^{\circ}\).

Step2: Substitute the values into the formula

Substitute \(a = 16\), \(b = 5\), and \(C = 61^{\circ}\) into the formula:
\(AB^{2}=16^{2}+5^{2}-2\times16\times5\times\cos(61^{\circ})\)
First, calculate \(16^{2}=256\), \(5^{2} = 25\), and \(2\times16\times5=160\). Then, \(\cos(61^{\circ})\approx0.4848\).
So, \(AB^{2}=256 + 25-160\times0.4848\)
\(AB^{2}=281-77.568\)
\(AB^{2}=203.432\)

Step3: Take the square root and round

Take the square root of \(203.432\): \(AB=\sqrt{203.432}\approx14.3\)

Answer:

\(14.3\)