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Question
question
in $\delta hij$, $h = 6.8$ cm, $i = 3.3$ cm and $j = 6.6$ cm. find the measure of $\angle j$ to the nearest 10th of a degree.
Step1: Recall the Law of Cosines
The Law of Cosines states that for a triangle with sides \(a\), \(b\), \(c\) and the angle \(C\) opposite side \(c\), \(c^{2}=a^{2}+b^{2}-2ab\cos(C)\). In \(\triangle HIJ\), to find \(\angle J\), the sides adjacent to \(\angle J\) are \(i = 3.3\) cm and \(h=6.8\) cm, and the side opposite \(\angle J\) is \(j = 6.6\) cm. So we can apply the Law of Cosines: \(j^{2}=h^{2}+i^{2}-2hi\cos(J)\).
Step2: Rearrange the formula to solve for \(\cos(J)\)
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Step3: Substitute the given values
Substitute \(h = 6.8\), \(i=3.3\), and \(j = 6.6\) into the formula:
\[
\]
Step4: Calculate the value of \(\cos(J)\)
\(\frac{13.57}{44.88}\approx0.3024\)
Step5: Find the angle \(J\)
Take the inverse cosine (arccos) of \(0.3024\) to find \(\angle J\): \(\angle J=\arccos(0.3024)\approx72.4^{\circ}\) (rounded to the nearest tenth of a degree)
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\(72.4^{\circ}\)