I've been given the following question:


The answer book states that the area of $\triangle ABC$ is $19.4cm^2$ and $\angle BDA$ is $52.16 ^ \circ$:

My understanding is that the $\sin$ law, when used to find angles, can provide 2 solutions. I found $\angle BDA$ to be $127.8 ^ \circ$ and the area of $\triangle ABC$ to be $13.47cm^2$. In the diagram the angle $\angle BDA$ is clearly obtuse, so I don't understand why I'm wrong.

Am I missing something or is there an error in the answer booklet?

Thank you.

  • $\begingroup$ Angle BDA (the angle between B and A measured from D) looks acute to me. $\endgroup$ – Paul Mar 25 '20 at 20:21
  • $\begingroup$ Do you know for sure that the figure is drawn to scale? $\endgroup$ – user170231 Mar 25 '20 at 20:28
  • $\begingroup$ Often diagrams are intentionally drawn with the acuteness/obtuseness not to scale. In this problem. Use law of sines to find the length of AD. drop a perpendicual above A and to the left of B, (call it R) to make triangle ARB with right angle ARB. The height of that triangle is $4.3 \sin 40$. The area of the big trianlge is $(8.6 + lengthAD)*(4.3 \sin 40)/2$. Make sure the sines are taken in degrees. Your calculation for the angle seems correct. $\endgroup$ – Mark Mar 25 '20 at 20:31
  • $\begingroup$ Looks to me like there are two solutions. $\endgroup$ – Doug M Mar 25 '20 at 20:37

First, be careful with text book diagrams. I always tell my students that unless it is specified to be drawn to scale (which is very rare) then the only information that can be taken as true in a diagram are the actual measurements provided.

Having said that, you are correct. There are two possible angles for $\angle BDA$
enter image description here

My guess is that, since the Sine Law gives the smaller of the two possible angles, the text book writers were not necessarily expecting you to go further than this (are there any worked examples prior to this exercise that might give a clue as to the writer's expectations?). Either that, or they haven't provided enough info in the diagram to make one or the other answer the 'correct' answer, or, whoever wrote the answer booklet didn't go far enough.

  • $\begingroup$ Ok. In future, I'll just assume the smaller angle is the correct one, unless told otherwise. The textbook is written by a UK exam board. $\endgroup$ – MrRookie Mar 25 '20 at 22:42

By the cosine's theorem, we have: $$x^2+2\cdot4.3\cdot\cos(40°)x+4.3^2-3.5^2$$ and so: $$x=\frac{8.6\cos(40°)\pm\sqrt{4\cdot4.3^2\cos^2(40°)-4(4.3^2-3.5^2)}}{2}=5.45 \: \vee \: 1.14$$

The first solution (obtained with $+$) is acceptable, the second (obtained with $-$) not. The area of the triangle is: $$A=\frac{1}{2}\cdot4.3\cdot(8.6+x)\cdot\sin(40°)=\frac{1}{2}\cdot4.3\cdot\left(8.6+\frac{8.6\cos(40°)-\sqrt{4\cdot4.3^2\cos^2(40°)-4(4.3^2-3.5^2)}}{2}\right)\cdot\sin(40°)=13.47$$

  • $\begingroup$ Why did you choose one value of $x$ over the other? I see two solutions. $\endgroup$ – Doug M Mar 25 '20 at 20:38
  • $\begingroup$ @Doug M: see my edits. Sorry for the misunderstanding. $\endgroup$ – Matteo Mar 25 '20 at 20:40
  • $\begingroup$ If the solution with the negative sign is not acceptable, then why did you use it later on? I think both are acceptable, and this problem has two solutions. $\endgroup$ – Doug M Mar 26 '20 at 17:21

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