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What is the value of $x$ in the figure ? All the numbers in degree. note that : $$\hat{A}+\hat{B}+\hat{C}=180$$ does not help in this case !I am stuck on this . enter image description here

Thanks in advanced.

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    $x=20$ works OK – 2017-02-11
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    @DavidQuinn :yes , you are right , but what is the method to solve ? – 2017-02-11
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    What are the "red lines"? – 2017-02-11
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    @Test123 :Red lines are not special . – 2017-02-11
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    What does that mean though? Should they intersect at an interior point? If not necessarily then even $x=0$ works. You need to put some conditions. – 2017-02-11
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    @Test123 :They intersect at an interior point . – 2017-02-11
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    It's essentially because the angles $2X$ and $120-4X$ are equal in the same way as the other corresponding pairs are. – 2017-02-11
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    @DavidQuinn :why they are equal ? – 2017-02-11
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    There is nothing that says they are equal. See my comment below. – 2017-02-11

6 Answers 6

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The trigonometric version of Ceva's theorem gives us $$ \frac{\sin x}{\sin 30^{\circ}}\cdot\frac{\sin (120^{\circ}-4x)}{\sin x}\cdot \frac{\sin 30^{\circ}}{\sin (2x)} = 1 $$ Therefore one of $$ 120^{\circ}-4x = 2x $$ or $$ 120^{\circ}-4x = 180^{\circ}-2x $$ must hold. The first equation gives us $x=20^{\circ}$, the second one yields a negative solution.

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It is not enough to say that the red lines meet in the same point.

(1) If the red lines are orthogonal to the opposite sides then $20^{\circ}$ is the solution.

(2) If the red lines are angle bisectors then there is no solution; it is impossible that $X=30^{\circ}$ (See $B$) and at the same time $2X=30^{\circ}.$ (See $A$)

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    It is enough to say that the red lines meet in the same point. This enforces the solution to be $20°$. The orthogonality is a side effect. – 2017-02-12
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Note that $$A=2x+30$$ $$B=x+30$$ $$C=120-4x+x=120-3x$$ Adding them all up gives $$2x+30+x+30+120-3x=180+3x-3x=180°$$ So, whatever value of $x$ you choose, it will still satisfy the equality.
There are still restrictions, however, as all angles of a triangle must be between $0°$ and $180°$
We need $$0° < 2x+30 < 180°$$ $$0° < x+30 < 180°$$ $$0° < 120-3x < 180°$$ Rearranging each inequality, we get $$ -15° < x < 75°$$ $$-30° < x < 150°$$ $$-20° < x < 40°$$ We need $x$ to satisfy all of these restrictions, so we'll take the intersection of these intervals, or the greatest lower bound and the least upper bounds of the inequalities, to get $$-15° < x < 40°$$ as all possible solutions.

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    This is nice, but this solution does not take it into account that the red lines meet at the same point. – 2017-02-11
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    True. My solution assumes the red lines are insignificant (from the asker's comment). Since your answer sums up that point, I won't copy it here. +1 on yours. – 2017-02-11
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    Also note that $x$ is the measure of an angle, so it surely won't be negative. So you have $0 < x < 40$. – 2017-02-11
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    @zoli: Indeed, the given conditions including concurrency of the cevians are enough to give a unique solution; please see my answer. – 2017-02-12
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I've modified your image a little to add some new labels. enter image description here

Call the interior point where the three red lines meet $S$.

(I've hidden some algebra below.)

Since angles $ASR$ and $BSQ$ are congruent, we have $u = 180 - v$. Similarly, angles $BSP$ and $CSR$ are congruent, so $w = 180 - v$. Hence $u = w$.

Since angles $ASP$ and $QSC$ are congruent, we have $150 - v = 3x$.

Analyzing the sum of angles around the point $S$ gives a tautology. Analyzing the sum of angles in any of the triangles $ABQ$, $PCB$, or $ACP$ give the same equation $150 - v = 3x$.

Someone should be clever and find another relation between $x$, $u$, and $v$, and then the problem is solved uniquely.

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    @Reinhard Meier I think Ceva's theorem tells us that $|AP| = |QC|$ in this case. Does that help, somehow? – 2017-02-11
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    Another observation: Triangle $SBP$ is similar to $SCR$, and triangle $SBQ$ is similar to $SAR$. Does this mean that triangle $SAP$ is also similar to $SCQ$? If so, then we know $u=v=90$ and then $x=20$. But I don't know that there is a theorem here about these similar triangles. – 2017-02-12
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    See my answer for how to finish. Indeed there is a unique solution. – 2017-02-12
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Here is a hint. Thanks to wckronholm for the image!

enter image description here

Hint:

There are two obvious cyclic quadrilaterals. Using them you can prove that $u = v = w$, and so there is a third cyclic quadrilateral. But then you can prove that $\angle AQP = \angle ACP$, giving a fourth cyclic quadrilateral. And its other pair of equal angles gives you $x$ easily.

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Method: Just add up all the expressions you have in the angles and the sum must be 180 degrees.

Working from the top and around the triangle clockwise,

2x + 30 + x + ... (you fill in the rest) .. = 180 degrees.

Then use basic algebra to solve for x.

(To those who are downvoting I was trying to get the OP to actually write out an equation and some limitations on the variables. You know, the basic "show your work" paradigm. If they just say they can't do this and we hand out a complete answer, well, that short circuits the learning process.)

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    $2x +30+x+30+120-4x+x=180$ for any $x$. – 2017-02-11
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    @victoria :I tried ,but solve nothing ...! can you show your work ? – 2017-02-11
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    All the $x$'s disappear if "fill in the rest", and your "equation" becomes $180=180$. That doesn't tell us anything. – 2017-02-11
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    This is very very basic algebra. I was hoping you would show your work since that is the only way you will learn. The comment above gives the equation. Now how do yu solve an equation like that? – 2017-02-11
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    If you come to 180 = 180 or x = x or 0 = 0 etc., that is something called a *tautology* It is true for any value of x. As an equation, you could fill in any real number at all. However in a triangle there are certain limits. x cannot be negative, being a real measure, so we get x is greater than or equal to zero. 120 - 4x also cannot be negative so what restriction does that give on x? Each of the angles must be less than 180 degrees. Does that give us any more restrictions on x, or are they already covered above? – 2017-02-11
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    There are additional restrictions that can be derived from the fact that the red lines intersect at one point. Maybe Ceva's theorem can help here. I will give it a try ... – 2017-02-11