Isosceles right triangle problem?

Question

Given isosceles right triangle ABC (A is the right angle) and point O inside which OBC = 30 degrees and OCB = 15 degrees. Prove that triangle OAB is isosceles.

Answer 1

It is Isosceles the only way to firure it out is to draw it well thats what works for me

Answer 2

Using Michael M's 9 initial findings, I am proposing to use the Law of Sines instead.

If we can prove angle OAB is 15, which would equal angle ABO, then triangle OAB is isosceles:

Let angle AOC = b and angle AOB = w
Now b + w +135 = 360, therefore, b + w = 225 or w = 225 - b

For triangle AOC: f / sin 30 = g / sin b
For triangle AOB: f / sin 15 = g / sin w = g / sin (225 - b)

Solving for g/f in both equations and equating them, we have

sin b / sin 30 = sin (225 - b) / sin 15

(sin b)sin 15 / sin 30 = sin 225 cos b - cos 225 sin b

0.5176 sin b = - 0.7071 cos b + 0.7071 sin b

combine like terms of sin b:

0.1895 sin b = 0.7071 cos b

recognizing sin b / cos b = tan b, we have:

tan b = 0.7071 / 0.1895 = 3.7313

b = arctan 3.7313 = 75 deg

w = 225 - b = 150

Therefore angle OAB = 180 - w - 15 = 15.

Answer 3

I seem to get 30 as well with the aid of a calculator. This question is deceivingly difficult. I went angle chasing at first, but then had to try something else since that only gets you so far. I used a similar Law of Sines approach to get eqns like sin(20 + x) = 2cos(40)sin(x) or cot(x) = (2cos(40) - cos(20)) / sin(20). I wasn't able to make much progress trying to solve these by hand.

Answer 4

see man.
a rt angled isosceles triangle can have angles 90, 45, 45 only.
so OBA is 15 degrees.
OCA is 30 degrees.
now tell me how can't it be correct

Answer 5

Well, Michael's solving seems to be right, but it's still complex and trigonometric.
Is there a better geometrical method way without (co)sines laws, by using some additional perpendiculars?

Answer 6

you can't. the only right triangle that is isocsoles is one with two 45 degree angles and one 90

Answer 7

d = distance from B to O
e = distance from C to O
f = distance from A to O
g = distance from A to B which equals A to C
(g)sqrt(2) = distance from B to C
Triangle ABC must equal a 90, 45, 45 triangle.
Angle ABO = 15 degrees
Angle ACO = 30 degrees
Angle BOC = 135 degrees

If d = f, then triangle OAB is isosceles.

Using law of cosines, we can relate f, d and g
f^2=d^2+g^2-2(d)(g)cos(15)
Using law of sins, we can relate g and d
(g)sqrt(2)/sin(135) = d/sin(15)

Now show d = f
g = (d)sin(135) / (sin(15)sqrt(2))
f^2=d^2 + g^2-2(d)(g)cos(15)
f^2=d^2 + ((d)sin(135) / (sin(15)sqrt(2)))^2 - 2(d)((d)sin(135) / (sin(15)sqrt(2)))cos(15)
f^2=d^2 + (d^2)sin(135)^2 / (2sin(15)^2) - 2(d)((d)sin(135) / (sin(15)sqrt(2)))cos(15)
f^2=d^2 + (d^2)sin(135)^2 / (2sin(15)^2) - 2(d^2)cos(15)sin(135) / (sin(15)sqrt(2))
f^2=d^2 + (d^2)(0.5) / (2sin(15)^2) - 2(d^2)cos(15)sin(135) / (sin(15)sqrt(2))
f^2=d^2 + (d^2)(0.25) / sin(15)^2 - 2(d^2)cos(15)(sqrt(0.5)) / (sin(15)sqrt(2))
f^2=d^2 + (d^2)(0.25) / sin(15)^2 - (d^2)cos(15) / (sin(15)
f^2=d^2 + (d^2)(0.25) / sin(15)^2 - (d^2)cos(15)sin(15) / (sin(15)^2
f^2=d^2 + (d^2)(0.25) / sin(15)^2 - (d^2)(0.25) / (sin(15)^2
f^2=d^2
f = d

That was perhaps the hardest trig problem I’ve ever done.

Answer 8

Michael is right !

Answer 9

can't