Determine the measure of the angle of the planes $(NBC)$ and $(ABC)$

Question

the problem

Let the triangle $ABC$ with $\angle A=30, \angle B=15, AC=2a$ and $M$ be the midpoint of $AB$. At point M we construct the perpendicular to the plane of the triangle on which we take point $N$ so that $MN=\frac{a\sqrt{2}}{2}$. Determine the measure of the angle of the planes $(NBC)$ and $(ABC)$

my idea

the drawing

I let $MX \perp BC$ and we know that $MN \perp (ABC)$ so by the theorem of the 3 perpendicular wecan say that $NX \perp BC$

This means that the angle we are looking for is the angle $NXM$

From here I didn't know what to do I tried calculating MX by area and trigonometri like the thorem of the cosine but got to nothing useful

Hope one of you can help me! Thank you!

Answer 1

These are the steps:

• Sum of angles in a triangle will give you $\angle C$
• Use law of sines in $\triangle ABC$ to relate $2x$ and $2a$
• in $\triangle BMX$ calculate $MX$
• You know at this point two sides in the right angle triangle $MXN$, so you should be able to compute the angle.

Let me know if any of the steps is not clear.