Here's the picture(I know I suck at paint):

Any help would be appreciated.

Originally Posted by RuyHayabusa
Here's the picture(I know I suck at paint):

Any help would be appreciated.
Hello

1. I've modified your sketch a little bit (see attachment): I refer to the labels of my sketch.

2. Use Pythagorian theorem and Cosine rule: You'll get 4 equations:
• $d^2=7^2+y^2$
• $d^2=x^2+4^2-2\cdot 4 \cdot x \cdot \cos(60^\circ)$
• $s^2=4^2+y^2-2 \cdot 4 \cdot y \cdot \cos(150^\circ)$
• $s^2=x^2+7^2-2 \cdot 7 \cdot x \cdot \cos(60^\circ)$

3. With $\cos(60^\circ)=\frac12$ and $\cos(150^\circ)=-\frac12 \cdot \sqrt{3}$ you'll get rid of the trigonometric functions.

4. The calculations are really ugly und you'll get many opportunities to make mistakes. For conformation only:
$(x,y,d,s)=(10,3\sqrt{3},2\sqrt{19},\sqrt{79})$

Originally Posted by earboth
Hello

1. I've modified your sketch a little bit (see attachment): I refer to the labels of my sketch.

2. Use Pythagorian theorem and Cosine rule: You'll get 4 equations:
• $d^2=7^2+y^2$
• $d^2=x^2+4^2-2\cdot 4 \cdot x \cdot \cos(60^\circ)$
• $s^2=4^2+y^2-2 \cdot 4 \cdot y \cdot \cos(150^\circ)$
• $s^2=x^2+7^2-2 \cdot 7 \cdot x \cdot \cos(60^\circ)$

3. With $\cos(60^\circ)=\frac12$ and $\cos(150^\circ)=-\frac12 \cdot \sqrt{3}$ you'll get rid of the trigonometric functions.

4. The calculations are really ugly und you'll get many opportunities to make mistakes. For conformation only:
$(x,y,d,s)=(10,3\sqrt{3},2\sqrt{19},\sqrt{79})$
Thanks,but I can't go anywhere from d^2=y^2+49=x^2-4x+16 , s^2=x^2-7x+49 = y^2+4*sqrt(3)*y

Originally Posted by RuyHayabusa
Thanks,but I can't go anywhere from d^2=y^2+49=x^2-4x+16 , s^2=x^2-7x+49 = y^2+4*sqrt(3)*y

This quadrilateral is a truncated equilateral triangle and a geometric solution leads to x =10 .Truncated section has two parts. A 30-60 -90 triangle and a regular trapozoid