Question #90785
Solve triangle ABC given that:A=60° ,AB=12,2cm and BC=14,5.
1
Expert's answer
2019-06-13T08:35:48-0400



We can use The Law of Sines first to find angle C:


ABsinC=BCsinA,\frac{AB}{\sin{C}}=\frac{BC}{\sin{A}},




12.2sinC=14.523,\frac{12.2}{\sin{C}}=\frac{14.5 \cdot 2}{\sqrt{3}},C=arcsin613145.C=\arcsin{\frac{61\sqrt{3}}{145}}.



The other angle C might be

180°arcsin613145133.23°,180 \degree - \arcsin{\frac{61\sqrt{3}}{145}} \approx 133.23 \degree,

but this is impossible, since the sum of the angles A and C is more than 180°.


Use "the three angles add to 180°" to find angle B:


B=180°AC=120°C.B = 180\degree - A - C = 120\degree - C.



Now we can use The Law of Sines again to find AC:


ACsinB=BCsinA,\frac{AC}{\sin{B}}=\frac{BC}{\sin{A}},


sinB=sin(120°C)=sin120°cosCcos120°sinC=321(613145)2(12)613145=3290(9862+61),\sin{B} = \sin{(120\degree - C)} = \\ \sin{120\degree}\cos{C}-\cos{120\degree}\sin{C} = \\ \frac{\sqrt{3}}{2} \sqrt{1-\left(\frac{61\sqrt{3}}{145}\right)^2} - \left(-\frac{1}{2}\right) \cdot \frac{61\sqrt{3}}{145} = \frac{\sqrt{3}}{290}\left(\sqrt{9862}+61\right),


AC=14.523sinB=110(9862+61).AC = \frac{14.5 \cdot 2}{\sqrt{3}} \sin{B} = \frac{1}{10} \left(\sqrt{9862}+61\right).

Answer:

C=arcsin61314546.77°,C=\arcsin{\frac{61\sqrt{3}}{145}} \approx 46.77 \degree,B=120°C73.23°,B = 120\degree - C \approx 73.23 \degree,AC=110(9862+61)16.03.AC = \frac{1}{10} \left(\sqrt{9862}+61\right) \approx 16.03.


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