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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