How to find $X$ with these given values?

Question Image :

The given values are:

$$\angle ABD = \angle CAD = 30^{\circ}$$
$$BD = DC$$

And we need to find $$\angle ACB = X$$ , how to solve this problem?

I tried to draw a parallel like that but couldn't go further with it:

• What have you tried? Show your attempt. Commented Nov 13, 2023 at 21:34
• Nice problem. However, it would be much nicer if you include your attempt to solve it. This way the community knows where you're having trouble. Commented Nov 13, 2023 at 21:35
• HINT: Insight will follow if you notice that one element of the drawing can be seen in two distinct ways. That will allow expression of equations to solve. Commented Nov 13, 2023 at 21:52
• @Prem You're really trying to un-HNQ it are you? Commented Nov 14, 2023 at 19:20
• I am not sure what you are saying , @mathlander , about HNQ ? I did edit to include MATHJAX. I had no other intention.
– Prem
Commented Nov 15, 2023 at 3:56

Applying the law of sines:

$$\dfrac{a}{\sin(30^\circ) } = \dfrac{b}{\sin(120^\circ - x)}$$

and

$$\dfrac{a}{\sin(x)} = \dfrac{b}{\sin(30^\circ)}$$

Dividing these two out,

$$\dfrac{\sin(x)}{\sin(30^\circ)} = \dfrac{\sin(30^\circ)}{\sin(120^\circ - x) }$$

Cross multiplying,

$$\sin(30^\circ)^2 = \sin(x) \sin(120^\circ - x)$$

Now, we know that

$$\sin(A) \sin(B) = \dfrac{1}{2} (\cos(A - B) - \cos(A + B) )$$

Thus our equation becomes,

$$\dfrac{1}{4} = \dfrac{1}{2} ( \cos(2 x - 120^\circ) - \cos(120^\circ) )$$

So that

$$\dfrac{1}{2} = \cos(2 x - 120^\circ) + \dfrac{1}{2}$$

From which

$$\cos(2 x - 120^\circ) = 0$$

Hence,

$$2 x - 120^\circ = -90^\circ$$ or $$2 x - 120^\circ = 90^\circ$$

And therefore,

$$x = \dfrac{120^\circ - 90^\circ}{2}$$ or $$x = \dfrac{ 120^\circ + 90^\circ}{2}$$

So that,

$$x = 15^\circ$$ or $$x = 105^\circ$$

Both are acceptable solutions.

Here are the two possible triangles. The first is with $$x = 15^\circ$$, the second with $$x = 105^\circ$$.

• Are you sure 105 degrees is an 'acceptable solution' ? Commented Nov 14, 2023 at 9:18
• Yes. I'm pretty sure. I've plotted the triangle with $x = 105$ and it is valid triangle. Commented Nov 14, 2023 at 9:27

Let:

• $$\overline{AD} = R.$$

• $$\overline{BD} = S = \overline{CD}.$$

• $$\angle BAD = \alpha.$$

Applying the Law of Sines, you have that

$$\frac{1/2}{S} = \frac{\sin(x)}{R} ~~~~\text{and} ~~~~\frac{1/2}{R} = \frac{\sin(\alpha)}{S} = \frac{\sin\left(120 - x\right)}{S}.$$

Therefore:

$$\displaystyle R = 2S\sin(x) \implies \frac{1}{2}S = 2S\sin(x) \times \sin(120 - x) \implies$$

$$\displaystyle \frac{1}{4} = \sin(x) \times \sin(120 - x) \implies$$

$$\displaystyle \frac{1}{4} = \sin(x) \times \left[\sin(120)\cos(x) - \sin(x)\cos(120)\right] \implies$$

$$\displaystyle \frac{1}{4} = \sin(x) \times \left[\frac{\sqrt{3}}{2}\cos(x) + \sin(x)\frac{1}{2}\right] \implies$$

$$\displaystyle 1 = \left[ ~2\sqrt{3}\cos(x) + 2\sin(x) ~\right] \times \sin(x) \implies$$

$$\displaystyle 1 - 2\sin^2(x) = 2\sqrt{3} \times \sqrt{1 - \sin^2(x)} \times \sin(x) \implies$$

$$\displaystyle \left[ ~1 - 2\sin^2(x) ~\right]^2 = \left[ ~2\sqrt{3} \times \sqrt{1 - \sin^2(x)} \times \sin(x) ~\right]^2 \implies$$

$$\displaystyle 1 - 4\sin^2(x) + 4\sin^4(x) = 12[1 - \sin^2(x)]\sin^2(x) \implies$$

$$\displaystyle 1 - 4\sin^2(x) + 4\sin^4(x) = 12\sin^2(x) - 12\sin^4(x) \implies$$

$$16\sin^4(x) - 16\sin^2(x) + 1 = 0. \tag1$$

Setting $$~u = \sin^2(x)~$$ allows (1) above to be interpreted as a quadratic in $$~u.$$

Therefore,

$$16u^2 - 16u + 1 = 0 \implies$$

$$\sin^2(x) = u = \frac{1}{32} \left[ ~16 \pm \sqrt{256 - 64} ~\right] \implies$$

$$\sin^2(x) = \frac{1}{4} \left[ ~2 \pm \sqrt{3} \right]. \tag2$$

Using a calculator, you have that $$~x \in \{15^\circ, 75^\circ\}.~$$

Checking each of the candidate values of $$~x~$$ against the constraint that
$$\displaystyle \frac{1}{4} = \sin(x) \times \sin(120 - x)$$

indicates that $$~x = 15^\circ.$$

• It's a beautiful solution, thank you so much!! Commented Nov 13, 2023 at 22:55
• @user2661923 Following your equation (2), note that $sin^2(x)=\frac{1}{2}[1-cos(2x)]=\frac{1}{4}(2\pm \sqrt{3})$. Then you can get $cos(2x)=\frac{\sqrt{3}}{2}$. This means $(2x)$ could be $\frac{\pi}{6}$, $\frac{7\pi}{6}$, $\frac{5\pi}{6}$, and $\frac{11\pi}{6}$ (radians). Only the first two are valid solutions. So finally you will get $x=15^\circ$ and $105^\circ$. Commented Nov 14, 2023 at 22:42
• @user295357 $~x = 105^\circ~$ is somewhat iffy, since the diagram seems to indicate that $~x~$ is an acute angle. Beyond that, I agree that your method of concluding the problem is more elegant than mine. However, from my perspective, once the calculator returned the candidate values of $~15^\circ, 75\circ,~$ one of which was eliminated by the initial constraint, there was no reason to pursue an alternate attack. Commented Nov 14, 2023 at 23:16
• Calculator is a convenient toy but can not be used to replace mathematic analysis. Sinusoidal function is a periodic function, and all valid solutions should be found out by strict math analysis. We shouldn't be limited by the diagram. I believe $x=105^\circ$ is one of valid solutions. Commented Nov 14, 2023 at 23:30

Guess and make special triangles.

$$\because \triangle ABC \sim \triangle DAC$$

$$\therefore AC^2 = BC*DC = 2DC^2$$

$$AC = \sqrt{2}DC$$

Draw $$CH\perp AD$$ and the intersection is H.

$$CH=\frac{1}{2}AC = \frac{\sqrt{2}}{2}DC$$

$$\therefore \triangle HDC$$ is an isosceles right triangle

$$\therefore \angle HCD=45^\circ$$

$$\because \angle ACH = 60^\circ$$

$$\therefore \angle x = 15^\circ$$

• What do you mean by "$\angle ACD = 60^\circ$" ? Does $\angle ACD =x$ ? By the way, how do you get "$CH=\frac{1}{2}AC$" ? Commented Nov 15, 2023 at 0:53
• @user295357 Sorry, it should be $\triangle ACH$, and △𝐴𝐶𝐻 is a special right triangle with a $30^\circ$ angle. So $𝐶𝐻=\frac{1}{2}𝐴𝐶$ Commented Nov 19, 2023 at 7:24
• Thanks for your reply and good job. By the way, after you get $AC=\sqrt{2}DC$, you can apply the sines law to get $\frac{sin 30^\circ}{DC}=\frac{sin \angle ADC}{AC}$. This will give you two answers for $\angle ADC$, that is, $\angle ADC=45^\circ$ and $\angle ADC=135^\circ$. Then finally you will get two answers for $\angle ACD$, that is, $\angle x=15^\circ$ and $\angle x=105^\circ$. Commented Nov 21, 2023 at 1:03
• Yes, thank you! Commented Dec 3, 2023 at 4:12

An alternate solution with minimal trigonometry: let $$B = (0, 0)$$; C = $$(0, 4)$$. Then $$(BA)$$ would be represented by an equation $$x - \sqrt{3}y=0$$. The set of points from which $$[DC]$$ is seen at $$30^\circ$$ would be $$(x-3)^2+(y-\sqrt{3})^2=4 | y>0$$ (easy to see from drawing right triangles with a $$30^\circ$$ angle).

Intersections are: $$(x, y) = (3-\sqrt{3}, \sqrt{3}-1), (3+\sqrt{3}, \sqrt{3}+1)$$. In other words, $$\tan \angle ACB = \frac{y}{4-x} = 2 - \sqrt{3}, -2 - \sqrt{3}$$. Checking known angles gives $$15^\circ$$, $$90^\circ+15^\circ = 105^\circ$$.