Question

Given a triangle with b = 4, c = 10, and A = 84º, what is the length of a? Round to the nearest tenth.

Find the area of the triangle with a = 11.8, b = 12.6, c = 14.8. Round to the nearest tenth.

Answer 1

(1) This is based on Law of Cosines:
\[c^2=a^2+b^2-2ab \cos(C)\]
In this case, we want a, so let's setup the law of cosines to solve for a:
\[a^2=b^2+c^2-2bc \cos(A)\]
Substituting in values we have:
\[a=\sqrt{(4)^2+(10)^2-2(4)(10)\cos(84)}\]\[a=\sqrt{116-80\cos(84)}\]
Therefore:
\[a=10.4\]
(2) This is based on Heron's Formula:
\[Area=\sqrt{p(p-a)(p-b)(p-c)}\]Where:
\[p=\frac{ a+b+c }{ 2 }\]
Rewriting Heron's Formula gives us:
\[Area=\sqrt{(\frac{ a+b+c }{ 2 })(\frac{ a+b+c }{ 2 }-a)(\frac{ a+b+c }{ 2 }-b)(\frac{ a+b+c }{ 2 }-c)}\]
Solving a+b+c / 2 we get:
\[p=\frac{ a+b+c }{ 2 }=\frac{ 11.8+12.6+14.8 }{ 2 }=19.6\]
Substituting in values we have:
\[Area= \sqrt{(19.6)(19.6-11.8)(19.6-12.6)(19.6-14.8)}\]
Therefore:
\[Area = 71.7\]