# Find the angle

#### razemsoft21

Can anyone help me finding the angle ADB ?

#### DenisB

Hint: if BC=1, then AB = SIN(70) / SIN(50) and AC = SIN(60) / SIN(50)

Is trigonometry allowed?

1 person

#### razemsoft21

Hint: if BC=1, then AB = SIN(70) / SIN(50) and AC = SIN(60) / SIN(50)

Is trigonometry allowed?
Sure everything is allowed.

#### DenisB

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!!

1 person

#### razemsoft21

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.
elegant non-trigonometric solution that mentioned above.

#### DenisB

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...

#### Idea

$$\displaystyle \frac{\sin u}{a}=\frac{\sin 130}{f}$$

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

$$\displaystyle \frac{\sin 50}{b}=\frac{\sin 70}{a}$$

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

#### Idea

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