In the diagram to the right, what is the value of x?A. \(1+\sqrt 2\)
B. \(1+\sqrt 3\)
C. \(2\sqrt 2\)
D. \(\sqrt 2 + \sqrt 3\)
E. \(2\sqrt 3\)
Check below:

Triangle BDC is a 30°-60°-90° right triangle. Now, in a right triangle where the angles are 30°, 60°, and 90° the sides are always in the ratio \(1 : \sqrt{3}: 2\), (
the leg opposite 30° (BD) corresponds with \(1\), the leg opposite 60° (DC) corresponds with \(\sqrt{3}\), and the hypotenuse \(BC=x\) corresponds with 2). Thus, since \(BC=x\) then \(BD=\frac{x}{2}\) and \(DC=\frac{x\sqrt{3}}{2}\).
Also notice that \(AD=AC-DC=x-\frac{x\sqrt{3}}{2}\).
Next, from Pythagoras theorem \(AB^2=BD^2+AD^2\);
\(2=(\frac{x}{2})^2+(x-\frac{x\sqrt{3}}{2})^2\);
\(2=x^2(\frac{1}{4}+1-2*\frac{\sqrt{3}}{2}+\frac{3}{4})\);
\(2=x^2(2-\sqrt{3})\);
\(x^2=\frac{2}{2-\sqrt{3}}\);
Rationalize: \(x^2=\frac{2*(2+\sqrt{3})}{(2-\sqrt{3})(2+\sqrt{3})}=4+2\sqrt{3}\);
\(x^2=4+2\sqrt{3}=(\sqrt{3})^2+2\sqrt{3}+1=(\sqrt{3}+1)^2\);
\(x=\sqrt{3}+1\).
Answer: B.
Hope it's clear.
Attachment:
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