# Math Help - basic geometry

1. ## basic geometry

Please see the attachment. Could anyone tell me how to determine the value of angle x? Thanks!

2. ## Re: basic geometry

Is there any more information given? Like are any of the segments parallel?

3. ## Re: basic geometry

sorry, no other info given...does that mean this problem isn't possible?

4. ## Re: basic geometry

Hi,
Your problem does have a solution. Unfortunately, I can't figure out how to do it geometrically. The following uses trigonometry to find x. The answer turns out to be very close to 16 degrees. This makes me think there is some geometric solution, but I am at a loss as to what it might be.

Thanks!

6. ## Re: basic geometry

Originally Posted by johng
Hi,
Your problem does have a solution. Unfortunately, I can't figure out how to do it geometrically. The following uses trigonometry to find x. The answer turns out to be very close to 16 degrees. This makes me think there is some geometric solution, but I am at a loss as to what it might be.

Looks like the angle is exactly 16 degrees.
Based on the fact that

$\sin \left(82^{\circ }\right)-\sin \left(38^{\circ }\right)=\sin \left(22^{\circ }\right)$

x + 22 = 38

Therefore
x = 16

7. ## Re: basic geometry

Originally Posted by Idea
Looks like the angle is exactly 16 degrees.
Based on the fact that

$\sin \left(82^{\circ }\right)-\sin \left(38^{\circ }\right)=\sin \left(22^{\circ }\right)$

x + 22 = 38

Therefore
x = 16
A more geometric proof of x + 22 = 38

Construct a right triangle EDI where I = 90 degrees, DI parallel to BC, angle EDI = 22 degrees

8. ## Re: basic geometry

Idea,
Could you explain to me how you get angle EDI is 22 degrees without knowing that angle x is 16 degrees? Of course angle ADI is 38.

9. ## Re: basic geometry

Originally Posted by johng
Idea,
Could you explain to me how you get angle EDI is 22 degrees without knowing that angle x is 16 degrees? Of course angle ADI is 38.

Construct EK perpendicular to BC, K on BC and as in your previous post a=BD=DE=EC

$\sin \left(82^{\circ }\right)=\text{EK}/a$

$\sin \left(38^{\circ }\right)=\text{IK}/a$

$\sin (\text{angleEDI})=\text{EI}/a=(\text{EK}-\text{IK})/a=\sin \left(82^{\circ }\right)-\sin \left(38^{\circ }\right)=\sin \left(22^{\circ }\right)$

10. ## Re: basic geometry

Hi Idea,
Thanks for your better proof of the original OP's problem. I'm still thinking about a purely geometric solution. The following is a diagram of your proof:

11. ## Re: basic geometry

Originally Posted by johng
Hi Idea,
Thanks for your better proof of the original OP's problem. I'm still thinking about a purely geometric solution. The following is a diagram of your proof:

Maybe if we can find a purely geometric proof that angle BCD = 30 degrees, it would then easily follow that angle x = 16

12. ## Re: basic geometry

Originally Posted by Idea
Maybe if we can find a purely geometric proof that angle BCD = 30 degrees, it would then easily follow that angle x = 16
Proof:
Extend side BA with AF = AC so that angle BFC = 30.

Show that triangles BCD and BFC are similar. Since they have an angle in common, this amounts to showing that BC*BC=BD*BF

This in turn follows from

$\text{DE}^2=\text{AD}^2+\text{AE}^2-\text{AD}.\text{AE}$

and

$\text{BC}^2=\text{AB}^2+\text{AC}^2-\text{AB}.\text{AC}$

(use Pythagoras' theorem)

Therefore angle BCD = angle BFC = 30 degrees