# solution-verification| calculate a trigonometric function of the angle of the plane (ABC) with the plane $\alpha$

the problem

The triangle ABC has vertex A in a plane $$\alpha$$ and is projected onto this plane according to the isosceles right triangle AMN, with $$AM=MN=a\sqrt{2}$$ M being the projection of B and N the projection of C. If $$BM=a\sqrt{3}, CN=2a\sqrt{3}$$ calculate a trigonometric function of the angle of the plane (ABC) with the plane $$\alpha$$, analyzing all cases. pOSSIBLE

the idea

As you can see I intrsected MN with BC in X and the intersection of the 2 planes would be XA.

By the reciprocal of the midline we get that M is the midpoint of XN, so $$XN=2a\sqrt{2}$$

in triangle XAN, MA= median and height, so traingle XAN is issoscel with $$XA=AN=2a$$ and by the reciprocal of the theorem of pytaghora we get that $$AN\perp XA$$

We can calculate BC in the rectangular trapeze MNCB, $$BC=a\sqrt{5}$$

By the reciprocal of the midline we get that B is the midpoint of XC, so $$XC=2a\sqrt{5}$$

$$AX=2a, ac=4a$$ and by the reciprocal of the theorem of Pythagoras we get $$CA\perp XA$$

Soo the angle between those planes if $$\angle CAN=60$$ by the reciprocal of the theorem of the angle of 30...so we can easily calculate a trigonometric function of it

I think that my idea is correct, but the phrase from the problem ,,analyzing all cases possible" its making me doubt a bit, because I didn't analyze any cases.

Hope one of you can help me! Thanks!

• Not quite sure I follow your description. $N$ is the projection of $B$ and $M$ is the projection of $C$? Anyway, I think possible cases are: 1) B and C are on the same side of the projection plane and 2) they are separated by the plane. Commented May 13 at 19:29
• @Vasili I edite my post...sorry for the misspelling .... Also, i dont really understand the other case...Can you please add an answer if you can and explain this case and how I should solve it, because I have no idea! Thank you! Commented May 14 at 7:21
• Wait. Is this really "solution-verfication" or something else? If it is indeed solution-veritication, then give the specific step you want checked. If it is not, then simply say 'here is my question, this is my attempt, can you answer/successfully complete my attempt'. Keep in mind that that this question/answer is not just for you but for whoever else may look at this.
– Mike
Commented May 15 at 19:44

It seems that you are assuming that $$X$$ is not between $$M$$ and $$N$$.

I think it is possible that $$X$$ is between $$M$$ and $$N$$.

In the following, I'm going to write my solution.

Let $$\theta$$ be the angle of the plane $$ABC$$ with $$\alpha$$.

We use $$X$$ that you defined. ($$X$$ exists since $$BC$$ is not parallel to the intersection line of the plane $$ABC$$ with $$\alpha$$.)

Let $$M'$$ be a point on $$AX$$ such that $$BM'\perp AX$$ and $$MM'\perp AX$$.

Let $$N'$$ be a point on $$AX$$ such that $$BN'\perp AX$$ and $$NN'\perp AX$$.

Since $$\triangle{BMM'}$$ is a right triangle with $$\angle{BM'M}=\theta$$, we have $$MM'=\frac{a\sqrt 3}{\tan\theta}\tag1$$

Since $$\triangle{CNN'}$$ is a right triangle with $$\angle{CN'N}=\theta$$, we have $$NN'=\frac{2a\sqrt 3}{\tan\theta}\tag2$$

It follows from $$(1)(2)$$ that $$NN'=2MM'$$.

Case 1 : $$X$$ is not between $$M$$ and $$N$$.

Let $$D$$ be the midpoint of the line segment $$NN'$$.

Then we have $$A=N'$$ since $$DM=\sqrt{MN^2-DN^2}=\sqrt{MA^2-MM'^2}=AM'$$

So, $$\frac 12NN'=\frac{MN}{\sqrt 2}$$ implies $$\color{red}{\tan\theta=\sqrt 3}$$.

Case 2 : $$X$$ is between $$M$$ and $$N$$.

Since $$\triangle{AM'M},\triangle{MM'X},\triangle{NN'X},\triangle{AN'N}$$ are right triangles, we have

$$AM'=\sqrt{AM^2-MM'^2}=\sqrt{2a^2-MM'^2}$$ $$XM'=\sqrt{MX^2-MM'^2}=\sqrt{\frac{2a^2}{9}-MM'^2}$$ $$XN'=2XM'=2\sqrt{\frac{2a^2}{9}-MM'^2}$$ $$AN'=\sqrt{AN^2-NN'^2}=\sqrt{4a^2-4MM'^2}$$

Since $$AN'=AM'+XM'+XN'$$, we have $$\sqrt{4a^2-4MM'^2}=\sqrt{2a^2-MM'^2}+3\sqrt{\frac{2a^2}{9}-MM'^2}$$

Squaring the both sides and simplifying give $$MM'^2=\sqrt{(2a^2-MM'^2)\bigg(\frac{2a^2}{9}-MM'^2\bigg)}$$ Squaring the both sides, we get $$MM'=\frac{a}{\sqrt 5}$$ which implies $$\color{red}{\tan\theta=\sqrt{15}}$$.