How do I find line AB in this if ac is $6cm$, and bc is $14cm$? angle A is $59^\circ$, B is $55^\circ$, and C is $66^\circ$. (not to scale)
thanks in advance
$\begingroup$
$\endgroup$
6
-
$\begingroup$ It is between 8cm and 20cm unless there are any more restrictions. $\endgroup$– Hagen von EitzenCommented Jun 26, 2013 at 17:52
-
$\begingroup$ @amWhy, 'scalene' nature is one $\endgroup$– lab bhattacharjeeCommented Jun 26, 2013 at 17:52
-
$\begingroup$ What, @lab? I don't get what you're saying. $\endgroup$– amWhyCommented Jun 26, 2013 at 17:54
-
$\begingroup$ @labbhattacharjee Well, when adding the condition 'scalene', only 14cm is prohibited $\endgroup$– Hagen von EitzenCommented Jun 26, 2013 at 17:54
-
2$\begingroup$ Given two sides, you need one angle to define the triangle. If the angle is not between the sides, there may be two solutions. If you define two (or three) angles, you have overspecified the problem and there may not (as here) be a solution at all. $\endgroup$– Ross MillikanCommented Jun 26, 2013 at 18:16
|
Show 1 more comment
4 Answers
$\begingroup$
$\endgroup$
3
Hint: with the added angles given, use the Law of Sines to compute the length of $\overline{AB}$. Taking angle B to be $55^\circ$
For angles A, B, C of a triangle, with $a$ being the length of the side opposite angle A, and so on, we have the following equality of ratios: $$\frac{a}{\sin A} = \frac{b}{\sin B} = \frac {c}{\sin C}\tag{Law of Sines}$$
$$\frac {14}{\sin(59^\circ)} = \frac {|AB|}{\sin(66^\circ)} = \frac{6}{\sin(55^\circ)}$$
As you can see, the first and last fraction are not equal, so no such triangle can exist.
-
$\begingroup$ what is the law of sines? I'm 13. (BTW I know algebra so you can use it to show me.) $\endgroup$ Commented Jun 26, 2013 at 18:00
-
$\begingroup$ Now that added conditions have been imposed, there is no such triangle. $\endgroup$ Commented Jun 26, 2013 at 18:01
-
$\begingroup$ Let me know if knowing the law of sines relationship helps. It is extremely useful in cases where a triangle exists, and you know some angles, but not all, in combination with some sides, but not all...And it doesn't require that the triangle be a right triangle to use it!! I applies to all triangles, when sufficient (consistent) information exists. $\endgroup$– amWhyCommented Jun 26, 2013 at 18:10
$\begingroup$
$\endgroup$
Without further information, you cannot say. With no knowledge about the angles, there are infinitely many triangles that have the lengths of two legs as $6$ and $14$.
answered Jun 26, 2013 at 17:52
$\begingroup$
$\endgroup$
2
If $AB=x$ cm
we need
$x+14>6,$ which is always true as $x>0$
$ 14+6>x \implies 20>x$
$6+x>14\implies x>8$
So, $x$ can assume any values between $(8,20)$
answered Jun 26, 2013 at 17:54
-
$\begingroup$ I think you mean in the last line: x+6>14 so x>8 $\endgroup$ Commented Jun 26, 2013 at 17:55
-
$\begingroup$ @user45878, yes, I meant that.Thanks $\endgroup$ Commented Jun 26, 2013 at 17:58
$\begingroup$
$\endgroup$
1
From $\frac {AB}{\sin C}=\frac{BC}{\sin A}$ we get $AB=13.14$cm. From $\frac {AB}{\sin C}=\frac{AC}{\sin B}$ we get $AB=6.28$cm. Note that not only the results differ significantly, the second one even dos not obey the triangle inequality. Where did you get your numbers from?
answered Jun 26, 2013 at 18:03