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

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!

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.

• I tried applying the thorem of sines but I doesn't help because we don't know the value of sin 15...Correct me if I'm wrong Commented Jul 6 at 12:33
• Use half angle formula. You know $\cos 30^\circ$ Commented Jul 6 at 12:40
• Commented Jul 6 at 12:44
• Thank you so much for your help! I got that $MX=\frac{a\sqrt{2}}{2}$ so the angle has $45$. Is it right? Commented Jul 6 at 12:49
• Correct. You notice that you don't even need to calculate $\sin 15^\circ$. It cancels out. Commented Jul 6 at 12:55