Find the angle x

Mar 2017
33
0
Norway
I've been working on this problem for days without any luck.
Any hint will be really appreciated.
DESB.jpg
 

skeeter

Math Team
Jul 2011
3,360
1,850
Texas
Looking for a clever way to solve this using just geometry. Maybe someone else can come up with a way to do that.

Hope you are able to use trigonometry on this problem.

Let the three equal sides have length 1

triangle ABD law of sines ... $a = \dfrac{\sin(48)}{\sin(66)}$

triangle BCD law of cosines ... $b = \sqrt{a^2+1-a\sqrt{3}}$

triangle BCD law of cosines ... $\cos{\theta} = \dfrac{a^2+b^2-1}{2ab} \implies \theta = \arccos\left(\dfrac{a^2+b^2-1}{2ab}\right)$

$x = 66 + \theta$


Angle_x.jpg
 
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Mar 2017
33
0
Norway
Looking for a clever way to solve this using just geometry. Maybe someone else can come up with a way to do that.

Hope you are able to use trigonometry on this problem.

Let the three equal sides have length 1

triangle ABD law of sines ... $a = \dfrac{\sin(48)}{\sin(66)}$

triangle BCD law of cosines ... $b = \sqrt{a^2+1-a\sqrt{3}}$

triangle BCD law of cosines ... $\cos{\theta} = \dfrac{a^2+b^2-1}{2ab} \implies \theta = \arccos\left(\dfrac{a^2+b^2-1}{2ab}\right)$

$x = 66 + \theta$


View attachment 11001
Thank you very much for your solution, although I have started to learn trigonometry and it will takes time for me but as you said maybe someone can come up with a geometry solution. Thanks again.
 
Last edited:
Oct 2013
713
91
New York, USA
skeeter's = 66 + has a variable, but it should be easy to calculate from what skeeter did. BD (what skeeter called a) can be calculated from the law of sines or law of cosines because all the other sides and angles in triangle ABD are known, and it is about 0.813. The lengths of 1 are made up, but if you call them x, BD would be about 0.813x, and the calculations for the angles would be the same. Then CD (what skeeter called b) can be can be calculated from the law of cosines with BD, BC, and angle CBD known, and it is about 0.503. Then angle CDB can be calculated from the law of sines or law of cosines, and it is about 84 degrees, making x about 150 degrees. Angle C would have to be 66 degrees. However, then I did BD/sin(66) = BC/sin(84) to check my work, and they weren't equal. What did I do wrong?
 

skeeter

Math Team
Jul 2011
3,360
1,850
Texas
angle CBD is obtuse by the law of cosines, which is why I used it ... if you used the law of sines to calculate it, you found the acute angle of 84 when it should be 96.
 
Oct 2013
713
91
New York, USA
My calculation of 84 was for angle CDB. Did you mean CDB when you typed CBD? I'm assuming that when my calculator does sin^-1 (I think the -1 exponent is what you said arccos for), it find the smallest possible angle with that sine, which will alway be from 0 to 90. If angle CDB is 96, angle CDA is 162, and angle DCB is 54. I didn't get the law of cosines to work for triangle BCD, but I don't care.
 

skeeter

Math Team
Jul 2011
3,360
1,850
Texas
yes, CDB = 96
 
Mar 2017
33
0
Norway
Hint: the diagram below can be used for a geometrical solution.
View attachment 11010
How are we going to construct this diagram? If we first draw $BF$ parallel to $CD$ then how are we going to find the angles? And if we first draw triangle $EBF$ with those angles then how are we going to find the angles of triangle $ECD$?
 

skipjack

Forum Staff
Dec 2006
21,481
2,470
If it can be drawn with those angles, they must be correct, as only one correct diagram is possible. Note that angle ADE is determined by the remaining angles shown - it has to be $108^\circ$.