I have a parallelogram $ABCD$ with side length $AB=58.7 cm$ and $BC=65.8 cm$ and $\angle DAB=120$ degrees.
I need to find the possible value(s) of $\angle CAB$.
Attempt.
This seems like a standard question, but I can't seem to figure out whether the angle is ambiguous or not, and why cosine rule would only give one solution but sine rule would give two solutions.
Using cointerior angles we can deduce $\angle ABC=60$ degrees.
$$AC=\sqrt{AB^2+BC^2-2AB\times BC\times \cos(60)}\approx62.55$$
Now to find $\angle CAB$ I have the option of using sine rule or cosine rule.
$$\frac{\sin(\angle CAB)}{65.8}=\frac{\sin(60)}{62.6}$$
$$\angle CAB=65^{\circ}39' \text{ or } 114^{\circ}22'$$
But if I find $\angle CAB$ with cosine rule I get
$$\cos(\angle CAB)=\frac{58.7^2+AC^2-65.8^2}{2\times58.7\times AC}$$
$$\angle CAB=65^{\circ}38'$$
Are both solutions valid? Or is only the acute angle valid? How would you tell whether this is an ambiguous case or not?
Thank you!
Best Answer
Ambiguous Case for Sine Rule is when you are given a presentation with two sides and an angle that is NOT between the two sides. For your parallelogram, as you deduced that $60^\circ$ was between two sides of known length -- then there is no ambiguous case here and we only take the acute angle.
A more calculation based way of showing that we should only accept the acute angle is if we compute, by Sine Rule, the size of $\angle DAC := x$
We would obtain $$ \frac{ \sin{x} }{58.7}=\frac{\sin{60}}{AC}\Rightarrow x=54^\circ22' \text{ or } 125^\circ38' $$
But since $x+\theta=120^\circ$ we can only take the acute angled solutions!
Hope this helps.