Question

(1) Find MN to the nearest tenth (2) In triangle TUV below, find m Thanks!

Answer 1

Answer:

1. 64.0

2. 88°

Step-by-step explanation:

1. By the cosine rule,

$MN^2=NL^2+ML^2-2ML\cdot NL\cdot \cos\angle L,\\ \\MN^2=36^+47^2-2\cdot 36\cdot 47\cdot \cos 100^{\circ},\\ \\MN^2=1296+2209-3384\cos100^{\circ}\approx 4092.625433,\\ \\MN\approx 63.97\approx 64.0.$

2. Use the cosine rule,

$UV^2=UT^2+VT^2-2UT\cdot VT\cdot \angle T,\\ \\8.1^2=4.2^2+7.1^2-2\cdot 4.2\cdot 7.1\cdot \cos \angle T,\\ \\65.61=17.64+50.41-59.64\cos \angle T,\\ \\\cos \angle T=\dfrac{65.61-17.64-50.41}{-59.64}=\dfrac{2.44}{59.64}\approx 0.0409,\\ \\\angle T\approx 88^{\circ}.$

Answer 2

Answer:

(1) NM = 64

(2) m∠T = 87.7°

Step-by-step explanation:

(1) By using cos Rule

∵ (NM)² = (ML)² + (NL)² - 2(ML)(NL)cosL

∴ (NM)² = (47)² + (36)² - 2(47)(36)cos100

∴ (NM)² = 4092.625433

∴ NM = 63.97 ≅ 64

(2) By using cos Rule:

∵ $cosT=\frac{(TU)^{2}+(TV)^{2}-(UV)^{2}}{2(TU)(TV)}$

∴ $cosT=\frac{(4.2)^{2}+(7.1)^{2}-(8.1)^{2}}{2(4.2)(7.1)}=\frac{61}{1491}$

∴m∠T= $cos^{-1}(\frac{61}{1491})=87.6552$

∴ m∠T = 87.7°