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Is this question solvable?

In $\Delta ABC$, $AB = a-b$ and $BC = 2\sqrt{ab}$, then $\angle B$ is

(a) $\: 60^{\circ}$

(b) $\: 30^{\circ}$

(c) $\: 90^{\circ}$

(d) $\: 45^{\circ}$

Community
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Kun
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    There is still enough freedom to have various (if not all) possible angles. – Hagen von Eitzen Aug 29 '16 at 14:03
  • is this correct: $$BC=\sqrt{ab}$$? – Dr. Sonnhard Graubner Aug 29 '16 at 14:08
  • is there another condition? – Dr. Sonnhard Graubner Aug 29 '16 at 14:08
  • BC = 2 root of (ab) as per the question. No other condition is given. I was wondering if this can solved with the given information. – Kun Aug 29 '16 at 14:14
  • can we use $$AC=b$$? – Dr. Sonnhard Graubner Aug 29 '16 at 14:17
  • Can the angle B be derived if we use AC = b. There are no more conditions given in the question. So I think any value of AC can be assumed so long as the value of angle B can be derived. I do not know how to solve it. That why I asked. – Kun Aug 29 '16 at 14:22
  • As pointed out in an answer, if we use the usual assignment of letters to the lengths of the sides then $a = BC$ and $b = AC$, so the problem statement says $AB = BC - AC$, which implies all the given choices for $\angle B$ are wrong. But if we do not assume $a = BC$ and $b = AC$ then we don't know enough about $a$ and $b$ to rule out any of the choices for $\angle B$. I think either there are clues surrounding this problem that you did not realize should be included, or the problem is a mistake. – David K Aug 29 '16 at 18:48

3 Answers3

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HINT:

What do we know from $\triangle\text{ABC}$:

  • $$\angle\text{A}+\angle\text{B}+\angle\text{C}=180^\circ=\pi\space\text{radians}$$
  • For the length $\text{AB}$: $$\text{AB}=\text{a}-\text{b}$$
  • For the length $\text{BC}$: $$\text{BC}=2\cdot\sqrt{\text{a}\cdot\text{b}}$$

Using the law of sinus and the law of cosines:

  • $$\frac{\text{BC}}{\sin\left(\angle\text{A}\right)}=\frac{\text{AC}}{\sin\left(\angle\text{B}\right)}=\frac{\text{AB}}{\sin\left(\angle\text{C}\right)}$$
  • $$ \begin{cases} \text{BC}^2=\text{AC}^2+\text{AB}^2-2\cdot\text{AC}\cdot\text{AB}\cdot\cos\left(\angle\text{A}\right)\\ \text{AC}^2=\text{BC}^2+\text{AB}^2-2\cdot\text{BC}\cdot\text{AB}\cdot\cos\left(\angle\text{B}\right)\\ \text{AB}^2=\text{BC}^2+\text{AC}^2-2\cdot\text{BC}\cdot\text{AC}\cdot\cos\left(\angle\text{C}\right) \end{cases} $$

So, using your information (and $\left(2\cdot\sqrt{\text{a}\cdot\text{b}}\right)^2=4\cdot\text{a}\cdot\text{b}$):

  • $$\frac{2\cdot\sqrt{\text{a}\cdot\text{b}}}{\sin\left(\angle\text{A}\right)}=\frac{\text{AC}}{\sin\left(\angle\text{B}\right)}=\frac{\text{a}-\text{b}}{\sin\left(\angle\text{C}\right)}$$
  • $$ \begin{cases} 4\cdot\text{a}\cdot\text{b}=\text{AC}^2+\left(\text{a}-\text{b}\right)^2-2\cdot\text{AC}\cdot\left(\text{a}-\text{b}\right)\cdot\cos\left(\angle\text{A}\right)\\ \text{AC}^2=4\cdot\text{a}\cdot\text{b}+\left(\text{a}-\text{b}\right)^2-2\cdot\left(2\cdot\sqrt{\text{a}\cdot\text{b}}\right)\cdot\left(\text{a}-\text{b}\right)\cdot\cos\left(\angle\text{B}\right)\\ \left(\text{a}-\text{b}\right)^2=4\cdot\text{a}\cdot\text{b}+\text{AC}^2-2\cdot\left(2\cdot\sqrt{\text{a}\cdot\text{b}}\right)\cdot\text{AC}\cdot\cos\left(\angle\text{C}\right) \end{cases} $$
Jan Eerland
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Setting $c=AB=a-b$ gives a degenerate triangle. If there is no relationship between the sides and $a$,$b$,$c$ others than stated any solution is possible. There is a lack of information.

H. H. Rugh
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Using @JanEerland ‘s suggestion, $BC = k \sin A$, $AC = k \sin B$, and $AB = k \sin C$; for some $k \ne 0$.

Substituting in the cosine law (wrt B), we get

$$\sin^2 B = \sin^2 A + \sin^2 C – 2 \sin A \sin C \cos B$$

(There might be others) but one solution of it is $B = 90^0$ and $A$ is then complement to $C$.

Note that, up to this point, the given $AB = a - b$ and $BC = ...$ have not been used. They probably are used to find $AC = … = a + b$ by Pythagoras theorem, and for the checking of triangle inequality (which maybe or maybe not necessary). This is not difficult to do with the help of $a – b \gt 0$.

However, the notation used in the question is very misleading. This is because according to our common naming convention, for $\triangle ABC$, $BC = a$, $AC = b$ etc. Such confusion can be found in various comments.

Mick
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