Question

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

Answer 1

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`.