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Question 1194000: in quadrilateral ABCD with diagonal BD, angle ABD is greater than angle CBD, AB = CD = BD = 25 and AD = 22, Determine the possible values of BC.
Answer by math_tutor2020(3817)   (Show Source):
You can put this solution on YOUR website!
For geometry problems like this, I strongly recommend a visual approach. Usually I would do a static drawing, but I'll try out a dynamic approach.
This is because one point is to move around freely.
Here is an interactive GeoGebra applet to play around with. Move point C around to note various possible lengths of BC. Make sure that angle CBD (red) is smaller than angle ABD (blue).
https://www.geogebra.org/m/wchywswb
You may need to click on the center if it says "press here to start the applet".
The button icon may be in the shape of a curved arrow.
Let me know if you have any technical difficulties running the applet.
I'll create a snapshot to show on here
In this current configuration, it's clear that the red angle is larger than the blue one.
This means BC = 24.55 isn't allowed.
However, a drawing like this helps us see what's going on.
Move point C to whichever location you like and create a snapshot based on it so you can get practice.
Focus on triangle ABD.
We have these side lengths
AB = 25
BD = 25
AD = 22
We can use the law of cosines to find angle B in blue to be roughly 52.21 degrees.
Here are the steps to see how we get that angle value
a^2 = b^2 + c^2 - 2*b*c*cos(A)
(AD)^2 = (AB)^2 + (BD)^2 - 2*(AB)*(BD)*cos(angle ABD)
(22)^2 = (25)^2 + (25)^2 - 2*(25)*(25)*cos(angle ABD)
484 = 1250 - 1250*cos(angle ABD)
-1250cos(angle ABD) = 484 - 1250
-1250cos(angle ABD) = -766
cos(angle ABD) = -766/(-1250)
cos(angle ABD) = 0.6128
angle ABD = arccos(0.6128)
angle ABD = 52.2077622746798
angle ABD = 52.21
The notation "arccos" refers to "arccosine". This is the same as inverse cosine denoted as 
Now we want angle ABD to be larger than angle CBD.
In other words, we want CBD to be smaller than ABD.
The possible angle measures of angle CBD would be
0 < angle CBD < 52.21
Turn your attention to triangle CBD.
We'll use the law of cosines again
Plug in 52.2077622746798 for angle CBD and let's compute how long side BC can be based on that.
I'll let x = length of BC
a^2 = b^2 + c^2 - 2*b*c*cos(A)
(CD)^2 = (BC)^2 + (BD)^2 - 2*(BC)*(BD)*cos(angle CBD)
(25)^2 = x^2 + (25)^2 - 2*x*(25)*cos(52.2077622746798)
625 = x^2 + 625 - 30.64x
x^2 - 30.64x = 0
x(x - 30.64) = 0
x = 0 or x-30.64 = 0
x = 0 or x = 30.64
Ignore x = 0 because BC = 0 isn't possible.
If BC = 30.64 approximately, then it leads to angle CBD being roughly 52.2077622746798 degrees.
If BC is anything smaller, then angle ABD won't be larger than angle CBD.
I recommend you trying to get BC to be 30.64 (or close to it) using the interactive GeoGebra app I posted above.
So effectively, BC = 30.64 is like the "floor" or the lowest we can go. We can't actually reach this value or else the red and blue angles would be equal.
The upper ceiling value is BC = 25+25 = 50 since we simply add sides BD = 25 and DC = 25 together.
We can't reach BC = 50 itself, so BC < 50 is what we're after.
Therefore, 30.64 < BC < 50 describes the set of all possible lengths for side BC where angle ABD is larger than angle CBD.
Feel free to use more decimal precision if your teacher requires it.
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