How do you solve the #triangle FGH# given #G=80^circ, H=40^circ, g=14#?

1 Answer
Dec 21, 2016

Triangle is #f=12.311#, #g=14#, #h=9.1381#, #F=60^o#, #G=80^o# and #H=40^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 three sides of triangle are #a#, #b# and #c# and angles opposite to them are #A#, #B# and #C#.

Here, we are given #DeltaFGH# and #g=14#, #/_G=80^o# and #H=40^o#. It is apparent that #/_F=180^o-80^o-40^o=60^o##

We can use sine formula to get side #h#. Using ^^Law of sines**

#f/(sin60^o)=14/(sin80^o)=h/(sin40^o)# and hence

#h=14/(sin80^o) xx sin40^o=14/0.9848xx0.6428=9.1381#

and #f=14/(sin80^o)xxsin60^o=14/0.9848xx0.866=12.311#

Hence, triangle is #f=12.311#, #g=14#, #h=9.1381#, #F=60^o#, #G=80^o# and #H=40^o#.