geometrytrigonometry
Is it possible to find the sides of a parallelogram knowing the length of its short diagonal and internal angles? The short diagonal measures 80 cm and its internal angles are 70° and 110°.
I tried law of sines but I need the angles on each rectangles, also I tried law of cosines but I couldn't. Obviously each internal triangle has angles 70°, $\alpha$, $\beta$ with $\alpha + \beta=110°$. But too many solutions, how can I pick the right one?
Fix one side, say segment $AB$, then draw (one arc of) the circle $c$ which consists of those points from where $AB$ is seen in the given angle. So that the centre of the parallelogramma has to be on $c$. Then enlarge $c$ to its double from center $A$, getting arc of circle $c'$ which must contain a $3$rd vertex of the parallelogramma, so finally just draw the circle centered at $B$ with the length of the other side.
Actually, you can't determine the area with just this information. We can determine the major base and the height, but we can't determine the minor base.
Here's how we determine the major base and the height:
Major Base: We use the law of sines to get the equation $$\frac{|MP|}{\sin(\angle MOP)}=\frac{|OM|}{\sin(\angle P)}$$ And we have everything we need there except $\angle MOP$, which we can calculate from $\angle MOP+\angle OMP+\angle P=180$
Height: We just note that $$\sin(\angle OMP)=\frac{h}{|OM|}$$
However, we cannot find the area because we can change the size of the minor base without changing any of our fixed values (which are $\angle P$, $\angle OMP$, and the length |OM|)
Here are some pictures showing this:
Best Answer
Not possible. A unique parallelogram cannot be constructed because other triangle of same subtended angle ( Euclid's thm ) can be found. All parallelograms have the same corner angles but different minimum distances between pair of parallel sides.
The sketch is not drawn to proper given angle size, the included angle is more towards 90 deg rather than 70 deg given.