# How to find the angle X of a triangle

Given a triangle whose vertex $A$ is bisecting the base (median), I have to find the angle $X$. From $\Delta$AHD we can get AH = HD. And it will make 45 degree at vertex A.

From $\Delta$ABH,

Let's Consider angle at B is '$\theta$'

tan $\theta$ = AH/BH

From here, I am not able to think forward.

Assumption: if somehow i can prove that BH = HD, then $\Delta$ABH will also be an isosceles triangle. Therefore total angle at vertex A will be 45 + 45 = 90 degree.

Possible Second Approach:

Assuming angle at B is 'Y'

From $\Delta$ABD

X + Y + 45 = 180

X + Y = 135 ----------- (i)

From $\Delta$ABC

(X + 15) + Y + 30 = 180

X + Y = 135 ------------(ii)

So, from above two equations I am not able to find X and Y (Since, Both are same)

Assumption: If I can get another linear equation in terms of X and Y. By solving those equation, I can get angle at 'A'

• Find the $2$ values of vertex $B$ and equate them. – For the love of maths Aug 23 '18 at 17:01
• Look at $\Delta ABD$ and $\Delta ABC$ – Entrepreneur Aug 23 '18 at 17:10
• @Entrepreneur Could you please detail? – Amit.D Aug 23 '18 at 17:23
• @MohammadZuhairKhan How those two value can be found? – Amit.D Aug 23 '18 at 17:40
• Hint: Drop $AH$ perpendicular to $CB$ extended. $AHD$ is right isosceles, and $CHA$ is a right triangle with a $30$ degree angle. – Lozenges Aug 23 '18 at 18:34

## 2 Answers

$\require{cancel}$

## Using completely trigonometry

Suppose $AC=q$, let us solve for all the sides in terms of $q$: $$\frac{\sin(135^\circ)}{q}=\frac{\sin(30^\circ)}{AD}\implies AD=\frac{q}{\sqrt2}$$ $$\frac{\sin(15^\circ)}{CD}=\frac{\sin(135^\circ)}{q}\implies\,CD=\frac{q(\sqrt3-1)}{2}$$ Since $CD=BD$, then $BC=q(\sqrt3-1)$. Then we need to find $AB$: $$AB^2=q^2+\left(q\left(\sqrt3-1\right)\right)^2-2q^2(\sqrt3-1)\cos(30^\circ)\implies AB=q\sqrt{2-\sqrt{3}}$$ Then we just need to solve for $x$ in: $$\frac{\sin x}{\frac{q(\sqrt3-1)}2}=\frac{\sin 45^\circ}{q\sqrt{2-\sqrt3}}\implies \sin x=\frac{\sqrt3-1}{2\sqrt{4-2\sqrt3}}$$ The nested radical in the bottom looks ugly, but remember that the radical $\sqrt{a\pm b\sqrt{c}}=\sqrt d\pm\sqrt e$ if and only if $\sqrt{a^2- b^2c}\,\,$ is rational. Therefore, we have: $$\sin x=\frac{\cancel{\sqrt3-1}}{2\cancel{(\sqrt3-1)}}=\frac12 \implies$$ $$\bbox[10px, border: 2px black solid]{\therefore x=\arcsin{\frac12}=\frac\pi6=30^\circ}$$

Here is one solution, which mostly uses basic geometry but uses trigonometry at the end.

Here is an extension of your diagram: I dropped a perpendicular from point $A$ to side $BC$ and called the intersection $H$, as you did. Since we are free to choose our units, let's set length $AH$ to one.

Now let's find other segment lengths. Since $HD$ is the side of a $45-45-90$ triangle, its length is the same as the other side: one. Since $HC$ is the long side of a $30-60-90$ triangle and the other side is one, its length is $\sqrt 3$. Subtracting those two, we get $DC=\sqrt 3-1$. By assumption, $BD=DC$ so $BD=\sqrt 3-1$. Finally, we get $BH = BD-HD=(\sqrt 3 - 1)-1 = \sqrt 3-2$.

Oops! The calculated length of $BH$, which is $\sqrt 3-2$, is negative! So something is wrong with the diagram. We see that point $H$ must lie to the left of point $B$ and angle $ABC$ must be obtuse, not acute as the original diagram supposed. If we correct the diagram, we get: The other lengths stay the same, but now $HB$ is $2-\sqrt 3$.

I added angle $Y$ which is angle $BAH$, in the diagram. Using triangle $BAH$ we see that

$$\tan Y=2-\sqrt 3$$

There are multiple ways to see that this means

$$Y=15°$$

You can get this by using the angle difference formula for $\tan(45°-30°)$. This can also be checked by expanding $\tan(15°+30°)$ or $\tan(2\cdot 15°)$. I'll leave that to you.

Finally, by looking at the $45-45-90$ triangle $ADH$ we see that $X+Y=45°$. Since $Y=15°$ we get our final answer:

$$X=30°$$

One way to check this is to draw my second diagram in Geogebra, which confirms the values for $X$ and $Y$.