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**Topic**

An’s family owns a triangular piece of land. The length of the fence *MN* is 150 m, the length of the fence *MP* is 230 m. Angle between two fences *MN *and *MP* is 110^{0} (Figure 21)

a) How many square meters is the area of land that An’s family owns (round up to tenths)?

b) Fence length *NP* how many meters (round to tenths)?

**Solution method – See details**

Step 1: Use the area formula \(S = \frac{1}{2}MN.MP\sin M\) to calculate the area ∆*MNP*

Step 2: Use the cosine theorem to calculate the length *NP *

Step 3: Conclusion

**Detailed explanation**

a)** **\({S_{MNP}} = \frac{1}{2}MN.MP\sin M = \frac{1}{2}.150.230.\sin {110^0} \approx 16209.7\) ( m^{2})

So the area of land that An’s family owns is 16209.7 m .^{2}

b) Apply the law of cosines to*ABC* we have: \(N{P^2} = M{N^2} + M{P^2} – 2.MN.MP.\cos M\)

\( \Rightarrow NP = \sqrt {M{N^2} + M{P^2} – 2.MN.MP.\cos M} \)\( = \sqrt {{{150}^2} + { {230}^2} – 2,150,230.\cos {{110}^0}} \approx 314.6\)(m)

So the length of the fence *NP *is 314.6 m

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