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Given a Triangle ABC with 3 Angles ($A, B, C$) and 3 sides ($a, b, c$) and Median $ma$ (from $A$). like image below:

Triangle Image

Problem 1: If we know Angle $A$, length of median $ma$, and side $a$. Is it possible to find one more side ($b$ or $c$) or angle ($B$ or $C$) ?

Problem 2: If we know Angle $B$, length of median $ma$, and side $b$. Is it possible to find one more side ($a$ or $c$) or angle ($A$ or $B$) ?

Maadhav
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Incpt.Mobis
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  • I took a look to the triangle formula: https://en.wikipedia.org/wiki/Triangle and Median Formulas: https://en.wikipedia.org/wiki/Median_(geometry). However, All of them relating to a median and two sides. I only have one side and an angle... I also tried expanding the triangle to parallelogram but no clue.... – Incpt.Mobis Dec 02 '17 at 03:22
  • You'll have to be more specific than that. Which formulas do you think are applicable here? How do you relate an angle with its sides – Dylan Dec 02 '17 at 03:23
  • Sorry, I hit enter. So, the comment is popped up incompletely. – Incpt.Mobis Dec 02 '17 at 03:24

1 Answers1

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Think big. Instead of trying to find just one line or one angle, try to find all of them. Or at least relate as many as you can.

For problem one, you know two sides and an angle. The other unknowns are the sides $b$, $c$ and angles $B$, $C$. You'll need at most a system of 4 equations. The law of cosines is useful here.

Triangle ABC gives $$ a^2 = b^2 + c^2 + 2bc\cos A $$ Triangle ABM gives $$ m^2 = b^2 + \left(\frac{a}{2}\right)^2 + ab\cos B $$

Triangle ACM gives $$ m^2 = c^2 + \left(\frac{a}{2}\right)^2 + ac\cos C $$

And of course $$ A + B + C = 180^\circ $$

With $A$, $m$, $a$ known, you have 4 equations in $b,c, B, C$ which can be solved. You don't have to solve this, but it's clear that a solution exists for all 4 variables.

Problem 2 is very much the same, but with different unknown variables.

Dylan
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