Find the angles of triangle

Question

We have a triangle $ABC$ where $(AC=BC)$. Point $D$ lay on $AB$ so $AD:BD=3:4$ and $\angle BCD=2\angle ACD$. How to find $cos\left(\angle BAC\right)\: $/which is equal to $cos\left(\angle ABC\right)\:$/ and$\:cos\left(\angle ACB\right)$?
Any ideas? I would be grateful for any tips!

Answer 1

So I drew this and found there are a few tricks in this question and one thing I think is an error.

AC = BC so the triangle is isosceles, yes. Angle ABC = angle BAC because the triangle is isosceles and the cosines of those angles are of course equal.

But we have no info about angle ACB. If its cosine were equal to the others. we'd have three equal angles and an equilateral triangle. That seems to overdetermine the problem and seems to contradict the placement of D. So for the moment let's set angle ACB aside, to find later.

Then we place D on AB so that AD:BD = 3:4

The first thought is to place D between A and B so that AD = 3/7 AB and BD = 4/7 AD. [I suggest you mark your first diagram with AD = 3x and BD = 4x; then of course AB = 7x.]

A second potential positioning is to extend AB in the direction of A and place D outside the triangle in order D, A, B with DA = 3 AB so DB = DA + AB = 4AB. Both positionings are possible within the given constraints so far. Draw a second diagram and look at it. [I suggest you mark your second diagram with AB = w and AD = 3w.].

Now we are given angle BCD = 2 angle ACD.

In version #1 with D between A and B, this means angle BCD = 2y and angle ACD = y [I suggest you mark these on your first diagram] so angle ACD = 1/3 angle ACB and angle BCD = 2/3 angle ACB. In version #2 with D exterior to the triangle, this means angle BCA = angle ACD = 1/2 angle BCD. [I suggest you mark each of angle BCA and angle ACD = z on your second diagram.]

Both diagrams satisfy the constraints of the question so far. Just by looking at it and trying to draw it Version #2 looks more likely but either may or may not be possible.

Now we have two potential solutions diagrammed and we need to calculate some values. We are looking for cosines so the first temptation would be to calculate the cosine law and see what happens. But that is long and complicated. We may have to do it, but is something simpler available?

I see the sine law and lots of equal sides and other related sides, and some related angles.|

Sine Law: in a triangle PQR with side p ofpposite angle P, side q opposite angle Q, and side r opposite angle R, [Note differences between capitals and lower case!! And note I use names different from the triangles we already have to avoid confusion.]

sin(P)/p = sin(Q)/q = sin(R)/r

OR p/sin(P) = q/sin(Q) = r/sin(R)

Remember that supplementary angles (angles which add to a straight line) have equal sines.

For convenience label AC = BC = a, CD = t, Angle CAB = Angle CBA = u.

In version #1 label angle ADC = s and angle BDC = 180 degrees - s so they have the same sine.

In version #2 label angle ADC = v; we see angle CAD = 180 degrees - u so sin(CAD) = sin u.

Now in Version #1, in triangle ADC, sin(s)/a = sin(u)/t = sin(y)/3x In triangle BDC, sin(s)/a = sin(u)/t = sin(2y)/4x and in the original triangle ABC, sin(u)/a = sin(3y)/7

sin(y)/3x = sin(2y)/4x

Multiply by x = x

sin(y)/3 = sin (2y)/4

Multiply by 4 = 4 Apply the sine addition formula

4/3 sin(y) = 2 sin(y)cos(y)

Divide by 2 = 2

2/3 sin(y) = sin(y)cos(y)

Either sin(y) = 0 or cos(y) = 2/3 Sin(y) = 0 --> y = 0 which is not a real triangle because angles are non-zero.

If cos y = 2/3, we can calculate sin(y) and use summation formulas to get sine and cosines for 2y and the other angles. But a quick reality check first. Using a calculator, arcsin(2/3) is approximately 41.8 degrees so angle ACB would be approximately 3*41.8 = 125.4 degrees and the other two angles of the triangle would each be 1/2(180 - 125.4) = approximately 27.3 degrees. Yes, this is a feasible triangle.

You can finish this one with summation formulas and the ratios above.

I strongly recommend you also work out Version #2 the same way. It is a valuable thinking exercise. And on tests sometimes the answer is not the one you expect.