Find ∠A & ∠C in Triangle Geometry

In summary, the conversation discusses finding angles ∠A and ∠C of a triangle △ABC without using software. Angle ∠B can be easily calculated using the given information and the sum-to-product trigonometric formula. However, the other angles require more complex methods such as trigonometry and geometry of regular pentagons. Ultimately, it is determined that ∠A=12° and ∠C=72°, which matches with the results from using Geogebra.
  • #1
Lemoine
1
0
Let △ABC be a triangle. Let AD and CE be its internal bisectors, with D lying on BC and E lying on AB. Given that ∠CED=18° and ∠ADE=24°, how can I find angles ∠A and ∠C without aid of softwares? Angle ∠B is easy to calculate, as
∠CED+∠ADE=∠CAD+∠ACE=∠A+∠C2=180°−∠B2
42°=180°−∠B2
∠B=96°
The other angles are difficult to calculate. I made this construction on Geogebra and got ∠A=12° and ∠C=72°(exactly), but I'm not being able to find them by geometric constructions. A teacher suggested me to drop perpendiculars from E to AD and from D to CE. Let E′ be the reflection of E over AD and D′ be the reflection of D over CE. Both D′ and E′ lie on AC, as AD bisects EE′ and CE bisects DD′.

What I found:
∠CED′=18°
∠ADE′=24°
∠DEE′=∠DE′E=90°−24°=66°
∠EDD′=∠ED′D=90°−18°=72°
∠E′ED′=66°−2⋅18°=30°
∠D′DE′=72°−2⋅24°=24°
 
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  • #2
Lemoine said:
Let △ABC be a triangle. Let AD and CE be its internal bisectors, with D lying on BC and E lying on AB. Given that ∠CED=18° and ∠ADE=24°, how can I find angles ∠A and ∠C without aid of softwares? Angle ∠B is easy to calculate, as
∠CED+∠ADE=∠CAD+∠ACE=∠A+∠C2=180°−∠B2
42°=180°−∠B2
∠B=96°
The other angles are difficult to calculate. I made this construction on Geogebra and got ∠A=12° and ∠C=72°(exactly), but I'm not being able to find them by geometric constructions.
I agree that this seems difficult. The only way I could get at it was to use trigonometry.
[TIKZ][scale=1.75]
\coordinate [label=above:{$A$}] (A) at (-6,0) ;
\coordinate [label=right:{$B$}] (B) at (0,0) ;
\coordinate [label=right:{$C$}] (C) at (0.5,3) ;
\coordinate [label=below:{$E$}] (E) at (-2,0) ;
\coordinate [label=right:{$D$}] (D) at (0.25,1.5) ;
\draw (A) -- (B) -- (C) -- (A) -- (D) -- (E) -- (C) ;
\node at (-5.1,0.1) {$\alpha$} ;
\node at (-5.1,0.3) {$\alpha$} ;
\node at (0.3,2.5) {$\gamma$} ;
\node at (0,2.6) {$\gamma$} ;
\node at (-0.17,0.15) {$96^\circ$} ;
\node at (-0.3,1.25) {$24^\circ$} ;
\node at (-1.4,0.5) {$18^\circ$} ;
[/TIKZ]
Suppose that the angles at $A$ and $C$ are $2\alpha$ and $2\gamma$. By the sine rule in $\triangle ABD$, \(\displaystyle \frac{BD}{\sin\alpha} = \frac{AB}{\sin(\alpha+96^\circ)}.\)

By the sine rule in $\triangle BEC$, \(\displaystyle \frac{BE}{\sin\gamma} = \frac{BC}{\sin(\gamma+96^\circ)}.\)

Using the sine rule in $\triangle BDE$ and then in $\triangle ABC$, it follows that $$\frac{\sin(\alpha+24^\circ)}{\sin(\gamma+18^\circ)} = \frac{BD}{BE} = \frac{AB\sin\alpha \sin(\gamma+96^\circ)}{BC\sin\gamma \sin(\alpha+96^\circ)} = \frac{\sin(2\gamma)\sin\alpha \sin(\gamma+96^\circ)}{\sin(2\alpha)\sin\gamma \sin(\alpha+96^\circ)} = \frac{\cos\gamma\sin(\gamma+96^\circ)}{\cos\alpha \sin(\alpha+96^\circ)}.$$
Therefore \(\displaystyle \sin(\alpha+24^\circ)\cos\alpha \sin(\alpha+96^\circ) = \sin(\gamma+18^\circ) \cos\gamma\sin(\gamma+96^\circ).\) Since $\alpha + \gamma = 42^\circ$, we can write this as $$\sin(\alpha+24^\circ)\cos\alpha \sin(\alpha+96^\circ) = \sin(60^\circ - \alpha) \cos(42^\circ - \alpha)\sin(138^\circ - \alpha^\circ).$$ In principle, that equation should determine $\alpha$. In practice, it is not so easy. I found that by messing about with sum-to-product trig formulas, I could reduce it to $$\tan\alpha = \frac{2\cos66^\circ - \sin48^\circ}{\cos48^\circ}.$$ To make progress from there, I had to use the geometry of the regular pentagon.

[TIKZ][scale=0.75]
\coordinate [label=above: $C$] (C) at (90:5cm) ;
\coordinate [label=above right: $D$] (D) at (18:5cm) ;
\coordinate [label=above left: $B$] (B) at (162:5cm) ;
\coordinate [label=below: $A$] (A) at (234:5cm) ;
\coordinate [label=below: $E$] (E) at (306:5cm) ;
\coordinate [label=below: $P$] (P) at (-4.755,-4.045) ;
\coordinate [label=above right: $N$] (N) at (0,-4.045) ;
\draw (A) -- node
{$1$} (B) -- node[above left] {$1$} (C) -- node[above right] {$1$} (D) -- node
{$1$} (E) -- node[below] {$1$} (A) -- (P) -- (B) -- (D) ;
\draw (C) -- (N) ;
\draw (0.3,1.25) node {$K$} ;
\draw (-3.4,-3.75) node {$72^\circ$} ;
\draw (-0.35,4.3) node {$54^\circ$} ;
\draw (0.4,4.3) node {$54^\circ$} ;
[/TIKZ]

In that diagram, $CN$ and $BP$ are perpendicular to $AE$. If the sides of the pentagon have unit length then $AN = \frac12$. But $AN = BK - PA = \sin54^\circ - \cos72^\circ$. So $$\sin54^\circ - \cos72^\circ = \tfrac12,$$ $$\sin(48^\circ + 6^\circ) = \cos72^\circ + \cos60^\circ = 2\cos6^\circ \cos66^\circ,$$ $$\sin48^\circ\cos6^\circ + \cos48^\circ\sin6^\circ = 2\cos6^\circ \cos66^\circ,$$ $$\cos48^\circ\sin6^\circ = \cos6^\circ(2\cos66^\circ - \sin48^\circ),$$ $$\tan6^\circ = \frac{2\cos66^\circ - \sin48^\circ}{\cos48^\circ}.$$ Therefore $\tan\alpha = \tan6^\circ$, so that $\angle A = 12^\circ$ and $\angle C = 72^\circ$, as forecast by Geogebra.​
 
Last edited:

Related to Find ∠A & ∠C in Triangle Geometry

1. What is the sum of all angles in a triangle?

The sum of all angles in a triangle is always 180 degrees. This is known as the Triangle Sum Theorem.

2. How do you find ∠A and ∠C in a triangle?

To find ∠A and ∠C in a triangle, you can use the Law of Sines or the Law of Cosines. You can also use the Triangle Sum Theorem to find the missing angles.

3. What is the difference between acute, right, and obtuse angles?

An acute angle is less than 90 degrees, a right angle is exactly 90 degrees, and an obtuse angle is greater than 90 degrees. In a triangle, all three angles must add up to 180 degrees.

4. How do you label the sides and angles of a triangle?

In a triangle, the side opposite the largest angle is called the hypotenuse. The other two sides are called the legs. The angles are typically labeled as ∠A, ∠B, and ∠C, with the corresponding lowercase letters labeling the opposite sides (a, b, c).

5. Can a triangle have more than one right angle?

No, a triangle can only have one right angle. If a triangle has more than one right angle, it would no longer be a triangle, as the angles would not add up to 180 degrees.

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