Answer to Question #349686 in Trigonometry for monique

Question #349686

find the missing parts of the triangle given below with 3 identified parts?

1.A=70 degrees 54' B=79 degrees 6' and a=20m

2.B=36 degrees 10' a=21m and b=30m

3.B=60degrees a=50m and c=60m

4.a=20m b=30m and c=40m


1
Expert's answer
2022-06-15T16:27:27-0400

Solution:

1) "A=70^{0}54'=70.9^{0};"

"B=79^{0}6'=79.1^{0};"

"a=20m." Let's find angle "C:"

"C=180-79.1-70.9=30^{0};"

From the sine theorem find other sides of triangle.

"\\frac{a}{sinB}=\\frac{b}{sinC};"

"b=a\\frac{sinC}{sinB}=20\\times\\frac{sin30}{sin79.1}=10.18m;"

"c=a\\frac{sinA}{sinB}=20\\times\\frac{sin70.9}{sin79.1}=19.25m;"

2) "B=36^{0}10'=36.16^{0};"

"a=21m ; b=30m;"

"\\frac{a}{sinB}=\\frac{b}{sinC};"


"sinC=sinB\\times(\\frac{b}{a})=\\frac{30}{21}\\times sin36.16=0.84;"

"C=arcsin(0.84)=57.14^{0};"

"A=180-B-C=180-36.16-57.14=86.7^{0};"

"c=b\\frac{sinA}{sinC}=30\\times\\frac{sin86.7}{sin57.14}=35.65m."

3) "B=60^{0}; \\space a=50m; \\space c=60m;"

"sinC=\\frac{a}{b}sinB=\\frac{50}{60}sin60=0.72;"

"C=arcsin(0.72)=46.1^{0};"

"A=180-60-46.1^{0}=73.9;"

"b=c\\frac{sinA}{sinB}=60\\times\\frac{sin73.9}{sin60}=66.56m."

4) "a=20m;\\space b=30m;\\space c=40m;"

Let's use the cosine theorem:

"a^{2}=b^{2}+c^{2}-2bccosC;"

"cosC=\\frac{b^{2}+c^{2}-a^{2}}{2bc}=\\frac{900+1600-400}{2\\times30\\times40}=0.87;"

"C=arccos(0.87)=29.54^{0};"

From the sine theorem:

"sinA=\\frac{b}{a}sinC=\\frac{30}{20}\\times sin29.54=0.74;"

"A=arcsin(0.74)=47.73^{0};"

"B=180-29.54-47.73=102.73^{0}."

Answer:

1) "C=30^{0}; \\space b=10.18m;\\space c=19.25m."

2)

"C=57.14^{0}; \\space A=86.7^{0}; \\space c=35.65m."

3)

"A=73.9^{0}; \\space C=46.1^{0};\\space b=66.56m."

4)

"A=47.73^{0}; \\space B =102.72^{0};\\space C=29.54^{0}."


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