How do you solve the triangle given C=80^circ, a=9, b=2?

1 Answer
Shwetank Mauria
Dec 18, 2016

Triangle is a=9, b=2, c=8.874, A=87.17^o, B=12.83^o and C=80^o.

Explanation:

Solving a triangle means identifying length of all the three sides as well as measures of all three angles. This is generally done using Law of sines, which is a/sinA=b/sinB=c/sinC and Law of cosines, according to which b^2=a^2+c^2-2ac cosB, c^2=a^2+b^2-2abcosC and a^2=b^2+c^2-2bc cosA

Here, we are given a=9, b=2 and C=80^o.

We can use c^2=9^2+2^2-2xx9xx2xxcos80^o

= 81+4-36xx0.17365=85-6.2514=78.7486

Hence c=sqrt78.7486=8.874

Now using ^^Law of sines**

9/sinA=2/sinB=8.874/(sin80^o)=8.874/0.9848=9.011

Hence sinA=9/9.011=0.9988 and A=87.17^o

and sinB=2/9.011=0.222 and B=12.83^o

Hence, triangle is a=9, b=2, c=8.874, A=87.17^o, B=12.83^o and C=80^o.

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