Thread: Weird trig triangle question, can't work through

Solve for theta, the big triangle is an isosceles. I've tried working it through by applying sine rule to the two separate partitions but can't get anywhere with the giant mess of an equation i get. I've got it to simplify to this.
, where u=θ/2
but can't get any further. I know θ should equal 90. But can't find a method to work through it.

There isn't enough info to solve this but make theta 90 degrees and prove the rest of the triangle.Does it work?

actually there is enough info ...

let = leg length of the large isosceles triangle







... it's just that the system of 3 unknowns doesn't look like much fun to solve by hand.

Hi skeeter,
You worked it into two unknowns.

i've already worked it into an equation with one unknown.

I did this by applying sine rule to each partition of the triangle to find b in terms of x, then putting the two partitions b's equal to each other.



i think that was it.. XD

Code:

                A

            (1)

                     (x)
       
      D  (b-e)         
  (x-1)         E
                       (e)
C     F         H               B

Label triangle as above; angle CBD = 15 (given). Let AB=AC=x; so CD=x-1.
AH = height line, intersects BD at E; let BE = e, then DE = b-e (since BD = b).

So e = SQRT(6)/2 / SIN(75); then calculate EH (simple pythagorean theorem).

Draw DF perpendicular to BC; triangle BDF is similar to triangle BEH ....so keep going....

Originally Posted by bjhopper

Hi skeeter,
You worked it into two unknowns.

, , and ... not assuming

Here's a fully integer example if you want to "play/practice!":

Code:

                    A


              51
                   63                   
           D
              30            119
                    E
     68
                   42        70

C        56         H       56          B

Your "b" is DB (length 100), crossing height line AH at E: DE = 30, BE = 70.

So the givens would be AD = 51, EH = 42 and BC = 112 (or BH = 56).

Dear skeeter,
I have puzzled over this problem for a couple of days.Triangle is right isosceles with side lengths rad 3.I cannot see how your equations can result in one equation of x alone.Would appreciate your help.

Originally Posted by bjhopper

I have puzzled over this problem for a couple of days.Triangle is right isosceles with side lengths rad 3.I cannot see how your equations can result in one equation of x alone.Would appreciate your help.

Well, perhaps this will cause you to get some sleep!

Using my diagram from my last post: place CB on x-axis with C at origin;
let h = AH, so A(56,h).

Letting (x,y) = D's coordinates, use following equations:
BD's y-intercept is clearly 84; hence BD's equation is: y = (-3/4)x + 84
AC's equation is easier still, with points (0,0) and (56,h) : y = (h/56)x

So we now need to solve:
(56 - x)^2 + (h - y)^2 = 51^2

Getting x and y in terms of h:
x = 4704 / (h + 42)
y = 84h / (h + 42)

And solving leads to this quartic:
h^4 - 84h^3 + 2299h^2 - 481908h + 943740 = 0
Which has h = 105 as only "valid" solution.

And that checks out ok. Makes the equal sides AB and AC = 119 ; also DE = 30.

Surprising to me that this is not easier, with right triangles all over the place!

Btw, if Sir Skeeter comes up with a quadratic for this, I'll eat my hat!!

Hi Wilmer,
The original post showed a specific triangle and asked to solve for angle at the peak.I did not see how,. this could be done aand suggested to assume it to be 90 degrees. Working backward all angles and line segments checked.Skeeter showed an algebraic solution (in part).Can you show what I'm asking

Getting the height as I showed provides sufficient info. to calculate the angle.

I used an example; just change my lengths to the OP's; for instance, make AD = 1, CH=BH=sqrt(6)/2

Dear Wilmer,
I would like to. see a solution using the OP drawing givens. b is not agiven

NEWS BULLETIN: your intuitive 90 degrees was CORRECT!!

Let h = height of triangle; place the triangle on x-axis:
(I'm using "b" as half the triangle's base; FORGET the OP's "b"!)

Code:

                 A(b,h)

             (1)

          D(x,y)

                 E(b,a)

                

C(0,0)           H(b,0)          B(2b,0)

We know that b = sqrt(6)/2 = ~1.224745
Since angle EBH = 15, then a = bSIN(15)/SIN(75) = ~.328169

Equation of AC: y = (h/b)x [1]
Equation of BD: y = (-a/b)x + 2a [2]

So, using [1] and [2]:
x = 2ab / (a + h)
y = 2ah / (a + h)

So, using AD as hypotenuse, we have:
(b - x)^2 + (h - y)^2 = 1^2
Substituting for x and y leads to:
h^4 - (2a)h^3 + (a^2 + b^2 + 1)h^2 - (2ab^2 + 2a)h + a^2(b^2 - 1) = 0

Using values to accuracy .000000:
h^4 - .656338h^3 + .607695h^2 - 1.640847h + .053847 = 0

Thanks to Wolfram: h = 1.224745

Let e = AB: e = sqrt(b^2 + h^2)
So angle BAH = ASIN(b/e) = 45 degrees !!
(my degree of accuracy results in 44.999996991.....)

Plus Sir Skeeter's warning that this wouldn't be short/simple was also correct!

EDIT: Note that triangle ADB = 30-60-90, thus the OP's b (line BD) = 2.

Thanks very much Wilmer. I understand the solution although the quartic is over my head