Catch the angle

By the figure, considering the triangleDNC:

$\frac{DN}{sin(30)} = \frac{DC}{sin(X+10)} \Rightarrow \frac{DN}{DC} = \frac{sin(30)}{sin(X+10)}$

By the triangle DBC:

$\frac{BD}{sin(50)} = \frac{DC}{sin(80)} \Rightarrow \frac{DN}{DC} = \frac{sin(50)}{sin(80)}$ , as BD=DN

So

$\frac{sin(30)}{sin(X+10)} = \frac{sin(50)}{sin(80)} \Rightarrow\\ sin(X+10) = \frac{sin(30)sin(80)}{sin50} = \frac{sin(30)2sin(40)cos(40)}{cos(40)} = sin(40)$

Or

$X+10 = arcsin(sin(40)) = 40°$ or $140°$

So

$X = 30º$ or $X = 130°$

Since X is acute, thus $X = 30 ^ \circ$ .

Please see the attached figure wherein it has been shown that triangle DBN is equilateral. But how may one do this by Euclidean geometry alone? Sreejato or anyone else, any ideas?

MY SOLUTION

A non-calculator, non-guess solution.

Triangle

The triangle is an isosceles triangle. Let:

• $AB\, =\, AB\, =\, 1$
• $DB\, =\, DN\, =\, BC\, =\, x$
• $\theta \, =\, \angle AND$

In $\triangle ABC$ , by sine law,

$\frac { x }{ \sin { 20° } } \, =\, \frac { 1 }{ \sin { 80° } } \\ \Rightarrow \, x\, =\, \frac { \sin { 20° } }{ \sin { 80° } }$

$\,$

Now, in $\triangle AND$ ,

$\frac { \sin { \theta } }{ 1\, -\, x } \, =\, \frac { \sin { 20° } }{ x } \\ \Rightarrow \, \sin { \theta } \, =\, \frac { 1\, -\, x }{ x } \sin { 20° } \\ \Rightarrow \, \frac { 1\, -\, x }{ x } \sin { 20° } \, =\, \frac { 1\, -\, \frac { \sin { 20° } }{ \sin { 80° } } }{ \frac { \sin { 20° } }{ \sin { 80° } } } \sin { 20° } \, =\, \sin { 80° } \, -\, \sin { 20° }$

$\,$

Using the identity: $\sin { u\, -\, \sin { v } } \, =\, 2\cos { \frac { u\, +\, v }{ 2 } } \sin { \frac { u\, -\, v }{ 2 } }$ , we have

$\sin { \theta \, =\, } \sin { 80° } \, -\, \sin { 20° } \, =\, 2\cos { 50° } \sin { 30° } \, =\, \cos { 50°\, =\, \sin { 40° } } \\ \Rightarrow \, \theta \, =\, 40°$

$\,$

We know that $\angle CNL\, =\, 10°$ because $\triangle CNL$ is a right triangle.

$\theta \, +\, \angle DNL\, +\, \angle CNL\, =\, 180°\\ \angle DNL\, =\, 180°\, -\, \theta \, -\, \angle CNL\, =\, 180°\, -\, 40°\, -\, 10°\, =\, \boxed { 130° }$