Question

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

Answer 1

(1) NM = 64

(2) m∠T = 87.7°

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=((TU)^(2)+(TV)^(2)-(UV)^(2))/(2(TU)(TV))`

∴
`cosT=((4.2)^(2)+(7.1)^(2)-(8.1)^(2))/(2(4.2)(7.1))=(61)/(1491)`

∴m∠T=
`cos^(-1)((61)/(1491))=87.6552`

∴ m∠T = 87.7°

Answer 2

1. 64.0

2. 88°

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=(65.61-17.64-50.41)/(-59.64)=(2.44)/(59.64)\approx 0.0409,\\ \\\angle T\approx 88^(\circ).`