2 use the cosine rule to find the missing lengths.
(a)
(b)
(c)
(d)

Question

Accepted Answer

# Explanation:
## Step1: Recall the cosine rule
The cosine rule is $c^{2}=a^{2}+b^{2}-2ab\cos C$. For part (a), let $a = b = 8$ and $C=42^{\circ}$.
## Step2: Substitute values into the cosine rule
$$
\begin{align*}
a^{2}&=8^{2}+8^{2}-2\times8\times8\times\cos42^{\circ}\
a^{2}&=64 + 64-128\times0.7431\
a^{2}&=128-95.1168\
a^{2}&=32.8832\
a&=\sqrt{32.8832}\approx5.73
\end{align*}
$$
For part (b), let $b = 9$, $c = 3$ and $A = 110^{\circ}$.
$$
\begin{align*}
a^{2}&=9^{2}+3^{2}-2\times9\times3\times\cos110^{\circ}\
a^{2}&=81+9 - 54\times(-0.3420)\
a^{2}&=90+18.468\
a^{2}&=108.468\
a&=\sqrt{108.468}\approx10.41
\end{align*}
$$
For part (c), let $a = 4$, $c = 7$ and $B = 132^{\circ}$.
$$
\begin{align*}
b^{2}&=4^{2}+7^{2}-2\times4\times7\times\cos132^{\circ}\
b^{2}&=16 + 49-56\times(-0.6691)\
b^{2}&=65 + 37.47\
b^{2}&=102.47\
b&=\sqrt{102.47}\approx10.12
\end{align*}
$$
For part (d), let $a = 6$, $b = 7$ and $C = 37^{\circ}$.
$$
\begin{align*}
c^{2}&=6^{2}+7^{2}-2\times6\times7\times\cos37^{\circ}\
c^{2}&=36+49 - 84\times0.7986\
c^{2}&=85-67.0824\
c^{2}&=17.9176\
c&=\sqrt{17.9176}\approx4.23
\end{align*}
$$

# Answer:
(a) $a\approx5.73$ cm; (b) $a\approx10.41$ cm; (c) $b\approx10.12$ cm; (d) $c\approx4.23$ cm

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QUESTION IMAGE

Question

Explanation:

Step1: Recall the cosine rule

Step2: Substitute values into the cosine rule

Answer: