Find the angle

Apr 2009
113
8
Jerusalem - Israel
Can anyone help me finding the angle ADB ?

2.jpg
 
Feb 2015
2,255
510
Ottawa Ontario
Hint: if BC=1, then AB = SIN(70) / SIN(50) and AC = SIN(60) / SIN(50)

Is trigonometry allowed?
 
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Feb 2015
2,255
510
Ottawa Ontario
a=AB, b=BC, c=CD, d=DA, e=AC, f=BD, u=angleADB
Let a=1. Angles ADC and ABC are both = 60 degrees...see that?

4 steps:

1: calculate e using triangleABC and Sine Law

2: calculate d using triangleACD and Sine Law

3: calculate f using triangleABD and Cosine Law

4: calculate u using triangleABD and Sine Law.
At this point, you should be using: a/SIN(u) = f/SIN(130)

There's probably a more elegant non-trigonometric solution,
but I'm too lazy to give it a shot!!
 
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Apr 2009
113
8
Jerusalem - Israel
e = sin(60)/sin(70)
d = sin(40)/sin(70)
f = sqr( 1 + d^2 - 2d con(13) )
using calculator:
sin(u) = 0.5
u = 30 deg.

Thanks a lot ...

It should be solved without using calc.
And please if you can give me an idea about the
elegant non-trigonometric solution that mentioned above.
 
Feb 2015
2,255
510
Ottawa Ontario
What made me say that was the fact that 30 degrees was answer,
so angleBDC also 30 degrees, thus angleABD=20 and angleCBD=40.

Since these angles are of integer size, possibility there is a
non-trigonometric solution is higher.
I don't have one...sorry. Perhaps someone else will...
 
Jun 2013
1,151
614
Lebanon
\(\displaystyle \frac{\sin u}{a}=\frac{\sin 130}{f}\)

\(\displaystyle \frac{\sin 110}{f}=\frac{\sin (60-u)}{b}\)

\(\displaystyle \frac{\sin 50}{b}=\frac{\sin 70}{a}\)

We multiply these equations to get \(\displaystyle \sin u = \sin (60-u)\) so \(\displaystyle u=30\)

If you think of \(\displaystyle sin\) as a ratio then no trig has been used.

More interesting, note that the two angles \(\displaystyle (50,80)\) satisfy the relation \(\displaystyle 2x+y=180\)

Similarly for \(\displaystyle (70,40)\). Also \(\displaystyle 50+70=80+40=120\).

Any four angles of the form \(\displaystyle (x,180-2x,120-x,2x-60)\) with \(\displaystyle 30<x<90\) will result in angle \(\displaystyle u=30\)

degrees
 
Jun 2013
1,151
614
Lebanon
here is a geometric proof

DA meets CB at E, AB is the interior bisector of angle EAC, so EA/AC = EB/BC

CE is the exterior bisector of angle C in triangle ACD, so EA/ED = AC/CD

Comparing the two proportions we see that ED/CD = EB/BC

This means that DB bisects angle D