I need to calculate the missing elements of a triangle, knowing its area one side and the angle opposite the given side. The triangle is not a right angle triangle, nor is it equilateral or isosceles.
Can someone advise a formula?
[Math] How to solve a triangle knowing its area, one side and an opposite angle
areatriangles
Related Solutions
The first condition is already fairly restrictive. Since $\angle A = 60^\circ$, $\cos \angle A = \frac12$, so we have $$a^2 = b^2 + c^2 - bc$$ by the Law of Cosines, which is nontrivial to satisfy over the integers. (We take $a,b,c$ to be the lengths of the sides opposite $A,B,C$.)
The existence of point $D$ gives us another Diophantine constraint. Draw the altitude $BH$ from $B$ onto $AC$:
Since $\triangle BCD$ is isosceles, $H$ is the midpoint of $CD$, so $CH = 1$; by looking at the right triangle $\triangle BHC$, we get $\cos \angle C = \frac{CH}{BC} = \frac1a$. The Law of Cosines tells us that $$c^2 = a^2 + b^2 - 2b.$$
Substituting the first equation into the second yields $$c^2 = (b^2 + c^2 - bc) + b^2 - 2b \implies bc = 2b^2 - 2b \implies c=2(b-1),$$ and substituting this value of $c$ back into the first equation tells us that $$a^2 = b^2 + (2b-2)^2 - b(2b-2) \implies a^2 - 3(b-1)^2 = 1,$$ which is a Pell equation. The simplest solution to $x^2-3y^2=1$ is $x=2,y=1$, and all other solutions are obtained by taking the coefficients of $1$ and $\sqrt3$ in $(2 + \sqrt3)^k$ for some $k$.
If we use $x=2$ and $y=1$, then this gives us $a=b=c=2$, which is certainly a solution to the problem: in this case, $\triangle ABC$ is equilateral and $D=A$. This triangle has area $\sqrt 3$.
If you don't like this solution, because it's a somewhat degenerate case, then $(2+\sqrt3)^2 = 7+4\sqrt3$ yields $x=7$ and $y=4$ as the next solution. In this case, $a=7$, $b=5$, and $c=8$, which is the triangle I used for the diagram. By Heron's formula, $\triangle ABC$ has area $10 \sqrt 3$.
One of these, depending on whether you find the equilateral solution acceptable or not, is the solution with the smallest area, but we can find infinitely many triangles with integer sides that satisfy the conditions, by taking higher powers of $2+\sqrt 3$ to solve Pell's equation.
Note that if $a$ and $b$ are two sides of the triangle and $\gamma$ is the angle in between then its Area is $$\text{Area}=\frac{a\cdot b\cdot\sin(\gamma)}{2}\implies b=\frac{2\cdot \text{Area}}{a\cdot \sin(\gamma)}$$ and we find another side. For the third side use the Heron's formula or the Law of Cosines and finally find the remaining angles from the Law of Sines. Can you take it from here?
Best Answer
Treat the side as the chord of a circle in which it subtends the given angle - the extended sine rule gives the radius. Then use the area of the triangle to calculate the height and you can identify the correct point on the circle.
Let the side you are given be $a$, and take this as the base of the triangle, and the angle opposite be $A$.
The extended version of the sine rule tells us that $\cfrac a{\sin A}=2R$ where $R$ is the circumradius of the triangle, so you can find $R$.
Having found $R$ you take the perpendicular bisector of your side, and locate one of the two points (above/below) which is distance $R$ from the extremities of the side (vertices $B, C$) and construct the circle.
With side $a$ as the base you calculate the height from area $=\cfrac {ah}2$.
Then take a line $l$ parallel to side $a$ and distance $h$ from it (on the correct side for the angle you want - if you get the wrong side the angle will be $180^{\circ}-A$). If the triangle is possible this will cut the circle in two points (you will see a symmetry about the perpendicular bisector) unless $l$ happens to be a tangent, when you get a single point. One of these points can be taken as vertex $A$.
I have not advised a formula, but a method. Following the method and doing the algebra will give you a formula if you need one.