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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!

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    $\begingroup$ How does this triangle have $4$ vertices? $\endgroup$ – Peter Foreman Apr 8 at 22:30
  • $\begingroup$ Sorry, parallelogram, editted $\endgroup$ – user523384 Apr 8 at 23:05
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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.

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If $\theta$ is between $0^\circ$ and $90^\circ$ then

$$\sin(\theta) = \sin(180^\circ - \theta) > 0$$

and $$\cos(\theta) = -\cos(180^\circ - \theta) > 0$$

In words, sine can't tell the difference between and angle and its supplement and cosine can.

You found the value of $AC$ and then you used $\overline{AB}, \ \overline{BC}$, and $\angle B$ of $\triangle ABC$ to find $m\angle BAC$ and got two answers. That is not suprising since, by SSA, you know that you have an underdetermined triangle.

The next step should have been to use the law of cosines to find the measure of $\angle BAC$. By $SSS$, you know that you will find the correct value.

Added because of conversation in the comments.

Since you now know that $AB<AC<BC$, then $\angle ACB$ must be an acute angle. Use the law of sines to find its value, then use $m∠BAC=180^\circ - 60^\circ −m∠ACB$.

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  • $\begingroup$ Thanks for the reply! I'm not sure whether I understood your answer though. Are you showing why there is chance for ambiguity with $\sin$? In some cases, there are two valid solutions for a situation, and sometimes there isn't. In this case, which one would it be? $\endgroup$ – user523384 Apr 9 at 3:54
  • $\begingroup$ Cosine rule can only be applied when the conditions are exactly those where the ambiguous case no longer applies. $\endgroup$ – Hugh Entwistle Apr 9 at 23:43
  • $\begingroup$ @HughEntwistle - Yes. Knowing $\sin \angle BAC$ alone does not tell you the measure of $\angle BAC$. Since $AB < AC < BC$, you could have worked it out by finding the value of $m\angle BCA$, which must be an acute angle, and then $m\angle BAC = 120^\circ - m\angle BCA$. $\endgroup$ – steven gregory Apr 10 at 12:35

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