Oops, sorry, must have crashed – dark, cold, rain, silence around these country parts, lie down, zonk.
How would one arrive at 50? 🙂 Difficult for you to start listing the angles you found, time consuming. I found there was a barrier beyond which I could find no further angles. I guessed 50 but could not be sure.
If EBCD were a cyclic quadrilateral, alpha would be 50 degrees because ECD is 50 degrees. But that can’t be right because alpha+CBE would be 90 degrees; hence EC would be a diameter and CDE would also have to be 90 degrees…
PS. I’ve drawn the figure using AutoCAD and I now know the answer. But I can’t see any way of getting to it by geometry alone.
Exactly. This is what I was hoping someone would explain. One would need a rightangle somewhere perhaps.
First thing to remember is that the drawing above is NOT TO SCALE!
It is possible using geam/trig, but you need to add 2 more lines –
Draw a line from D to line BC, at an angle of 20 deg to line DC, to intersect BC at a point we’ll call G.
Draw in another new line EG.
Angle GDC was drawn as 20 deg. and angle DCG is 80 deg, so angle DGC is also 80 deg, which means that line DC = line DG (as triangle DCG is isosceles)
Similarly, angle DCE is 50 deg, angle CBE is 80 deg, so angle DEC is 50 and line DE = line DC (this triangle is also isosceles).
From this DE = DG, and since angle EDG is 60 deg, triangle EDG is equilateral.
And triangle GBD is isosceles with angle GBD = 40 deg.
And finally, triangle GBE is isosceles with angle 70 deg, so angle alpha is 70-40 = 30 deg.
Simple plodding but my adding up has turned to mush today. Calculate F by 180 – ( 60 + 50) = 70. As the lines cross the side opposite F is the same as F. That lets you calculate the other angles in the middle as (360 – 2F) / 2. = 110 You now have two angles at the right and left.
On the right you have 110 + 30 to give the angle in the corner by alpha 40
On the left you have 110 +20 to give the angle to complete that one as 50
A is 180 – ( 20 + 60 ) – (30 + 50) = 20
So now we have to go to lengths and use sins and cosigns to calculate it out.
So full length of left side = 1 Have no idea but we don’t need size to be right just angles.
Full length of right side = 1 as well as both angles = 80 at bottom
So need to calculate size of top component on left using sins so dist between A and B is 0.53209
Top of left using same calcs = 0.653
So we now calculate the angle in the top of B using cosigns as 110
So alpha = 180 – 110 – 40 = 30
That should be right.
Can’t go to lengths because the drawing is not to scale. It can’t be – just look at the 70 and 110.
Sure you can. Lengths can not be determined but can be used to calculate the angles as we have just demonstrated.
I agree with Lord T that alpha is 30 degrees. I have different working.
I am quite happy to use algebraic values for lengths, and more geometry, and cope with the diagram not being to scale. On this, I don’t really understand James’ objection (if it is an objection).
I’ll type up my working, which might take some time.
First, I calculated and wrote in all the angles that can be simply calculated from sum angles of a triangle and sum angles on one side of a straight line.
Triangles DAC, CDE and ABD are each isosceles – with (in each triangle) the centre vertexes being the unequal ones.
To simplify things, I define the side DC to have unit length (DC=1.0). This differs from Lord T’s definition of DA=AB=1.0.
The following two algebraic equalities are used.
(i) The Law of Sines is that, for any triangle (vertices PQR, angles pqr), the length of each side divided by the sine of the opposite angle. Thus PQ/sin(r) = QR/sin(p) = RP/sin(q).
(ii) The sine of a sum of angles is given by sin(p+q) = sin(p).cos(q) + cos(p).sin(q).
Length EC is 1.28558 which is given from the Law of Sines for triangle CDE: EC/sin(80) = 1.0/sin(50)
Length EB (first equation) is given from the Law of Sines for triangle BDE: EB/sin(20) = 1.0/sin(alpha)
Length EB (second equation) is given from the Law of Sines for triangle BCE: EB/sin(30) = EC/sin(alpha+40)
Solving simultaneously by eliminating EB and putting in the length EC gives: sin(20)/[sin(alpha).sin(30)] = 1.28558/sin(alpha+40).
This ‘simplifies’ to: sin(alpha+40)/sin(alpha) = 1.28558.sin(30)/sin(20).
Using the sine of sum of angles for sin(alpha+40) gives: [sin(alpha).cos(40)+cos(alpha).sin(40)] / sin(alpha) = 1.87939
Cancelling terms and rearranging gives: cotan(alpha) = [1.87939-cos(40)] / sin(40) = 1.73206
Taking the reciprocal gives: tan(alpha) = 0.5775; hence alpha = 29.99984 degrees.
Assuming a ‘simple’ answer and that there is a bit of arithmetic rounding to allow for, we have alpha = 30 degrees.
Hoping I have made no mistakes of copying or algebra.
Which makes Lord T correct and that we must never admit, I might lose money on it. 🙂
Let us give thanks and praise for the mathematically minded. I can make fine coffee. And hum a tune.
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