How to find $X$ with these given values?

Question

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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:

Answer 1

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 $x = 105^\circ$.

Answer 2

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 $~x = 15^\circ.$

Answer 3

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

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

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

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

Answer 4

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 $15^\circ$, $90^\circ+15^\circ = 105^\circ$.