Question

Find x. See attached picture.

Answer 1

Answer:

B. 22

Step-by-step explanation:

We know tan60=

$\sqrt{3}$

Let common side be x

Tan = opp/adj= sqrt(3)

$\frac{11 \sqrt{6} }{x} = \sqrt{3}$

Transposing we get :

$\frac{11 \sqrt{6} }{ \sqrt{3}} = x = 11 \sqrt{2}$

Now, we know that since the other triangle is a 45-45-90 triangle, we can state that the side adjacent to the given angle = 11 sqrt(2) as the two sides are equal due to the triangle being an isoceles triangle.

Solve for x by using Pythagorean theorem:

$\sqrt((11 \sqrt{2}^{2} + (11 \sqrt{2} )^{2} ) = x$

$= > \sqrt{2(11 \times 11 \times \sqrt{2} \times \sqrt{2}) } = x$

$= > \sqrt{484} = x = 22$

Hence answer is B. 22