In the Triangle shown in the figure what is the value of x ?

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by Birottam Dutta » Thu Jul 12, 2012 8:00 am
If you apply the cosine formula, we get,

(sqrt2)^2 = x^2 + x^2 - 2.(x)(x)cos 30 = 2x^2 - 2x^2 (sqrt3/2) = 2x^2 - (sqrt3)x^2

=> 2 = 2x^2 - (sqrt3)x^2

on solving for x^2, we get, x^2 = 2/(2-sqrt3)

Multiplying both numerator and denominator by (2 + sqrt 3), we get x^2 = 2(2+sqrt3)/(4-3)=2(2+sqrt 3)

= 4+ 2(sqrt3)

This can further be simplified as follows:

4+ 2(sqrt3) = 1 + (sqrt3)^2 + 2*1*(sqrt3) = {1+ (sqrt3)}^2

So, x^2 = {1+ (sqrt3)}^2

=> x = 1+ (sqrt3).

Hence, answer is B!

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by GMATGuruNY » Thu Jul 12, 2012 9:10 am
gmatter2012 wrote:In the Triangle shown in the figure what is the value of x ?
Since the figure includes a 30 degree angle, draw a 30-60-90 triangle:
Image

The sides of a 30-60-90 triangle are proportioned s : s√3 : 2s.
Thus, BD = x/2 and DC = (x√3)/2.

We can plug in the answers, which represent the value of x.
The figure indicates that BD<√2.
Eliminate C, D and E, which will yield too great a value for BD.
Since a 30-60-90 triangle includes a side of √3, the correct answer is likely to be B.

Answer choice B: 1 + √3
BD = x/2 = (1+√3)/2

DC = (x√3)/2 = (1+√3)(√3) / 2 = (√3+3)/2.

AD = x - DC = (1+√3) - (√3+3)/2 = (2 + 2√3 - √3 - 3) / 2 = (√3-1)/2.

For B to be correct, AD² + BD² = AB²:
(√3-1)² / 2² + (1+√3)² / 2² = (√2)²

(3 - 2√3 + 1)/4 + (1 + 2√3 + 3)/4 = 2

8/4 = 2

2=2.

Success!

The correct answer is B.
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