In the diagram, what is the value of x?

$$A.\ 1+\sqrt{2}$$
$$B.\ 1+\sqrt{3}$$
$$C.\ 2\sqrt{2}$$
$$D.\ \sqrt{2}+\sqrt{3}$$
$$E.\ 2\sqrt{3}$$

The OA is B.

I don't have clear this PS question. I appreciate if any expert explain it for me. Thank you so much.

Hi AAPL,
Let's take a look at your question.

We will solve this problem using Cosine law. Cosine law is
$$a^2=b^2+c^2-2bc\ Cos\ A$$
Where a, b and c are the side lengths of a triangle and A is the angle between the sides b and c.

Apply Cosine law on the triangle give, we get:
$$\left(\sqrt{2}\right)^2=x^2+x^2-2\left(x\right)\left(x\right)\ Cos\ 30$$
$$2=2x^2-2x^2\ Cos\ 30$$
$$1=x^2-x^2\ \frac{\left(\sqrt{3}\right)}{2}$$
$$1=x^2\left(1-\frac{\sqrt{3}}{2}\right)$$
$$1=x^2\left(\frac{2-\sqrt{3}}{2}\right)$$
$$x^2=\frac{2}{2-\sqrt{3}}$$

Let's rationalize the denominator to simplify it.
$$x^2=\frac{2}{2-\sqrt{3}}\times\frac{2+\sqrt{3}}{2+\sqrt{3}}$$
$$x^2=\frac{2\left(2+\sqrt{3}\right)}{\left(2\right)^2-\left(\sqrt{3}\right)^2}$$
$$x^2=\frac{2\left(2+\sqrt{3}\right)}{4-3}$$
$$x^2=\frac{2\left(2+\sqrt{3}\right)}{1}$$
$$x^2=4+2\sqrt{3}$$

We need to see if we can write the right hand side as a square of a number? Let's try writing it as a square.
$$x^2=1+3+2\sqrt{3}$$
$$x^2=\left(1\right)^2+\left(\sqrt{3}\right)^2+2\left(1\right)\sqrt{3}$$

Using formula a^2+b^2+2ab=(a+b)^2, we can write the above equation as:
$$x^2=\left(1+\sqrt{3}\right)^2$$
$$x=\left(1+\sqrt{3}\right)$$

Therefore, Option B is correct.

Hope it helps.
I am available if you'd like any follow up.

AAPL wrote:

Zoom here, https://postimg.cc/image/u12yw856d/

In the diagram, what is the value of x?

$$A.\ 1+\sqrt{2}$$
$$B.\ 1+\sqrt{3}$$
$$C.\ 2\sqrt{2}$$
$$D.\ \sqrt{2}+\sqrt{3}$$
$$E.\ 2\sqrt{3}$$

The GMAT does not expect knowledge of cosines.
This problem can be solved using two common triangles:
45-45-90 and 30-60-90.

Step 1: Divide the 30-75-75 triangle into common triangles


Step 2: Calculate the sides of 45-45-90 triangle ADC
In a 45-45-90 triangle, the sides are in the following ratio:
s : s : s√2.
Thus, sides of triangle ADC are as follows:
AD = 1, DC=1, AC = √2.


Step 3: Calculate the sides of 30-60-90 triangle DEC
In a 30-60-90 triangle, the sides are in the following ratio:
s : s√3 : 2s.
In triangle DEC, DC = s√3.
Since DC = 1, we get:
s√3 = 1
s = 1/√3.
Thus:
DE = s = 1/√3.
CE = 2s = 2/√3.


Step 5: Calculate the sides of isosceles triangle ABE
The figure for Step 4 implies the following
BE = AE = 1 + 1/√3.


6. Determine the value of x in isosceles triangle ABC
The figure for Step 5 indicates the following:
x = BC = 2/√3 + (1 + 1√3) = 1 + 3/√3 = 1 + √3.

The correct answer is B.