How to find value of x?

Question

Find the value of x.

Trail: Let consider the point between B and C is D. So $\angle ADB =120^\circ$ and $\angle DAB =15^\circ$. I believe that $\angle DAC =30^\circ$. But I am not sure. Please help me. Thanks in advance.

Answer 1

Apply the Sine Rule to $\Delta ADB$ and $\Delta ABC$ to find two expressions for $AB$. That lets you relate $\sin x$ to $\sin(x+45)$. Then solve for $\tan x$.

Answer 2

So from $\triangle ADB$ using sine rule, we get $$\frac{\sin 15 }{1}=\frac{\sin 120 }{AB}=\frac{\sin 45 }{AD}$$ Similarly from $\triangle ABC$ using sine rule, we get $$\frac{\sin (135-x) }{3}=\frac{\sin x }{AB}=\frac{\sin 45 }{AC}$$

So equating $AB$ from this two equations we get $$\frac{\sin (135-x) }{\sin x }=\frac{3\sin 15 }{\sin 120}$$

Then I'm stuck.

Answer 3

Let $h$ be the height from A to BC,

$$\frac{ h\cot45 - h\cot60}{h\cot x + h\cot60} = \frac12 $$ $$\cot x = 2\cot 45 -3 \cot 60=2-\sqrt3$$ $$ x = 75$$

Answer 4

One way or another (there are multiple straightforward though slightly tedious ways to do this), show that $AC = \sqrt 6.$

For example, drop a perpendicular from $A$ to $BC$; let the intersection with $BC$ be $H$. Then $\angle BAH=45^\circ$, $\angle DAH=60^\circ,$ and if the length of $AH$ is $h$ then $$1 = h(\tan(45^\circ)-\tan(60^\circ)) = \frac{3 - \sqrt3}3 h.$$ Hence $h = \frac12(3 + \sqrt3).$ But the length of $CH$ is $3-h = \frac12(3 - \sqrt3),$ and $(AC)^2 = h^2 + (3-h)^2 = 6.$

Then there are at least two quick ways to solve the problem.

One way is to observe that $\sqrt 6$ is the geometric mean of $CD=2$ and $BC=3.$ Therefore triangles $\triangle ABC$ and $\triangle DAC$ are similar with $\angle DAC = \angle ABC,$ and now you have two angles of $\triangle DAC$ and the third angle is easy to find.

Another way is to use the Law of Sines: $$ \frac{\sin(60^\circ)}{\sqrt6} = \frac{\sin(\angle DAC)}{2}. $$ You can solve for $\sin(\angle DAC),$ and then (as it happens in this question) it is easy to see what $\angle DAC$ must be.