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anemone
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In an equilateral triangle $ABC$, let $D$ be a point inside the triangle such that $\angle BAD=54^\circ$ and $\angle BCD=48^\circ$. Prove that $\angle DBA=42^\circ$.
Olinguito said:I think I’ve got it!
I showed that in my post above that
$$a\ =\ \frac{1-\sqrt3\tan18^\circ}{1-\tan18^\circ\tan54^\circ}.$$
But
$$\tan54^\circ\ =\ \frac1{\tan36^\circ}\ =\ \frac{1-\tan^218^\circ}{2\tan18^\circ}$$
$\implies\ \tan18^\circ\tan54^\circ\ =\ \dfrac{1-\tan^218^\circ}2$
$\implies\ 1-\tan18^\circ\tan54^\circ\ =\ \dfrac{1+\tan^218^\circ}2$
$\implies\ a\ =\ \dfrac{2(1-\sqrt3\tan18^\circ)}{1+\tan^218^\circ}$
$\implies\ 2-a\ =\ 2-\dfrac{2(1-\sqrt3\tan18^\circ)}{1+\tan^218^\circ}\ =\ \dfrac{2\tan18^\circ(\sqrt3+\tan18^\circ)}{1+\tan^218^\circ}.$
$\implies\ \dfrac a{2-a}\ =\ \dfrac{(1-\sqrt3\tan18^\circ)}{\sqrt3+\tan18^\circ}\cdot\dfrac1{\tan18^\circ}$.
Also
$$\tan{78^\circ}\ =\ \tan(60+18)^\circ\ =\ \frac{\sqrt3+\tan18^\circ}{1-\sqrt3\tan18^\circ}$$
and the quantity $\dfrac b{2-a}$ in my post above is supposed to be $\tan42^\circ$. So my entire solution reduces to proving the following:
$$\boxed{\tan42^\circ\ =\ \frac{\tan54^\circ}{\tan78^\circ\cdot\tan18^\circ}}.$$
It’s quite simple, really: just show that
$$\tan x^\circ\tan(60+x)^\circ\tan(60-x)^\circ\ =\ \tan(3x)^\circ.$$
Proof:
$$\begin{array}{rcl}\tan x^\circ\tan(60+x)^\circ\tan(60-x)^\circ &=& \tan x^\circ\cdot\dfrac{\sqrt3+\tan x^\circ}{1-\sqrt3\tan x^\circ}\cdot\dfrac{\sqrt3-\tan x^\circ}{1+\sqrt3\tan x^\circ} \\\\ {} &=& \dfrac{3\tan x^\circ-\tan^3x^\circ}{1-3\tan^2x^\circ} \\\\ {} &=& \tan(3x)^\circ.\end{array}$$
Now put $x=12$:
$$\tan12^\circ\tan72^\circ\tan48^\circ\ =\ \tan36^\circ.$$
Now use the fact that $\tan(90-\theta)^\circ=\dfrac1{\tan\theta^\circ}$:
$$\frac1{\tan78^\circ\tan18^\circ\tan42^\circ} =\ \frac1{\tan54^\circ}$$
$\implies\ \tan42^\circ\ =\ \dfrac{\tan54^\circ}{\tan78^\circ\cdot\tan18^\circ}$
– QED! (Clapping)
Dhamnekar Winod said:Hello,
It can be proved using sine and cosine theorem in a very short period of time and without unnecessary work.
F A
E
D C B
An equilateral triangle is a type of triangle where all three sides are equal in length and all three angles are also equal. This means that each angle in an equilateral triangle measures 60 degrees.
In order to prove that ∠DBA=42° in an equilateral triangle, we can use the fact that the sum of the angles in any triangle is 180 degrees. Since we know that an equilateral triangle has three equal angles, we can divide 180 degrees by 3 to get 60 degrees. Then, we can use the fact that the angles in a straight line add up to 180 degrees to find that ∠DBA must be 60 degrees - 18 degrees = 42 degrees.
No, the Pythagorean Theorem is used to find the length of the sides of a right triangle, not the angles. In an equilateral triangle, all three angles are equal and therefore the Pythagorean Theorem is not applicable.
Other methods that can be used to prove ∠DBA=42° in an equilateral triangle include using the properties of congruent triangles, the Law of Sines, or the Law of Cosines. However, using the fact that the sum of the angles in a triangle is 180 degrees is the most straightforward and efficient method.
No, in an equilateral triangle, all three angles are equal and therefore must measure 60 degrees each. This means that ∠DBA can only measure 60 degrees, and since we have proven that it is 18 degrees less than that, it must be 60 degrees - 18 degrees = 42 degrees.