10N

Simulare Constanța Evaluarea Națională Matematică 2023

28 februarie 2023
Subiect
Barem

Subiectul I

1.
5

Rezultatul calculului
40−2⋅32\def\arraystretch{1.5} \def,{{\char`,}} \def\Div{{\space\raisebox{-0.1em}{$\vdots$}\space}} {40 - 2 \cdot 3^2}
este:

A
4\def\arraystretch{1.5} \def,{{\char`,}} \def\Div{{\space\raisebox{-0.1em}{$\vdots$}\space}} {4}
B
22\def\arraystretch{1.5} \def,{{\char`,}} \def\Div{{\space\raisebox{-0.1em}{$\vdots$}\space}} {22}
C
342\def\arraystretch{1.5} \def,{{\char`,}} \def\Div{{\space\raisebox{-0.1em}{$\vdots$}\space}} {342}
D
28\def\arraystretch{1.5} \def,{{\char`,}} \def\Div{{\space\raisebox{-0.1em}{$\vdots$}\space}} {28}
2.
5

Numărul întreg
x\def\arraystretch{1.5} \def,{{\char`,}} \def\Div{{\space\raisebox{-0.1em}{$\vdots$}\space}} {x}
care verifică ecuația
2x−6=1\def\arraystretch{1.5} \def,{{\char`,}} \def\Div{{\space\raisebox{-0.1em}{$\vdots$}\space}} {2^{x-6} = 1}
este:

A
0\def\arraystretch{1.5} \def,{{\char`,}} \def\Div{{\space\raisebox{-0.1em}{$\vdots$}\space}} {0}
B
−6\def\arraystretch{1.5} \def,{{\char`,}} \def\Div{{\space\raisebox{-0.1em}{$\vdots$}\space}} {-6}
C
6\def\arraystretch{1.5} \def,{{\char`,}} \def\Div{{\space\raisebox{-0.1em}{$\vdots$}\space}} {6}
D
2\def\arraystretch{1.5} \def,{{\char`,}} \def\Div{{\space\raisebox{-0.1em}{$\vdots$}\space}} {2}
3.
5

Suma numerelor naturale prime, mai mici decât
10\def\arraystretch{1.5} \def,{{\char`,}} \def\Div{{\space\raisebox{-0.1em}{$\vdots$}\space}} {10}
, este:

A
25\def\arraystretch{1.5} \def,{{\char`,}} \def\Div{{\space\raisebox{-0.1em}{$\vdots$}\space}} {25}
B
18\def\arraystretch{1.5} \def,{{\char`,}} \def\Div{{\space\raisebox{-0.1em}{$\vdots$}\space}} {18}
C
17\def\arraystretch{1.5} \def,{{\char`,}} \def\Div{{\space\raisebox{-0.1em}{$\vdots$}\space}} {17}
D
12\def\arraystretch{1.5} \def,{{\char`,}} \def\Div{{\space\raisebox{-0.1em}{$\vdots$}\space}} {12}
4.
5

Ordinea crescătoare a numerelor
x=−35, y=−53 și z=−215\def\arraystretch{1.5} \def,{{\char`,}} \def\Div{{\space\raisebox{-0.1em}{$\vdots$}\space}} {x = -3\sqrt{5},\text{ }y = -5\sqrt{3} \text{ și } z = -2\sqrt{15}}
este:

A
y;z;x\def\arraystretch{1.5} \def,{{\char`,}} \def\Div{{\space\raisebox{-0.1em}{$\vdots$}\space}} {y; z; x}
B
x;y;z\def\arraystretch{1.5} \def,{{\char`,}} \def\Div{{\space\raisebox{-0.1em}{$\vdots$}\space}} {x; y; z}
C
x;z;y\def\arraystretch{1.5} \def,{{\char`,}} \def\Div{{\space\raisebox{-0.1em}{$\vdots$}\space}} {x; z; y}
D
z;y;x\def\arraystretch{1.5} \def,{{\char`,}} \def\Div{{\space\raisebox{-0.1em}{$\vdots$}\space}} {z; y; x}
5.
5

Calculând
15%\def\arraystretch{1.5} \def,{{\char`,}} \def\Div{{\space\raisebox{-0.1em}{$\vdots$}\space}} {15\%}
din
420\def\arraystretch{1.5} \def,{{\char`,}} \def\Div{{\space\raisebox{-0.1em}{$\vdots$}\space}} {420}
se obține:

A
60\def\arraystretch{1.5} \def,{{\char`,}} \def\Div{{\space\raisebox{-0.1em}{$\vdots$}\space}} {60}
B
42\def\arraystretch{1.5} \def,{{\char`,}} \def\Div{{\space\raisebox{-0.1em}{$\vdots$}\space}} {42}
C
62\def\arraystretch{1.5} \def,{{\char`,}} \def\Div{{\space\raisebox{-0.1em}{$\vdots$}\space}} {62}
D
63\def\arraystretch{1.5} \def,{{\char`,}} \def\Div{{\space\raisebox{-0.1em}{$\vdots$}\space}} {63}
6.
5

Andrei afirmă că valoarea absolută a numărului
−8,2\def\arraystretch{1.5} \def,{{\char`,}} \def\Div{{\space\raisebox{-0.1em}{$\vdots$}\space}} {-8,2}
este
8,2\def\arraystretch{1.5} \def,{{\char`,}} \def\Div{{\space\raisebox{-0.1em}{$\vdots$}\space}} {8,2}
. Afirmația lui Andrei este:

A
adevărată
B
falsă

Subiectul al II-lea

1.
5

În figura alăturată, punctele
A, B și C\def\arraystretch{1.5} \def,{{\char`,}} \def\Div{{\space\raisebox{-0.1em}{$\vdots$}\space}} {A,\text{ }B \text{ și } C}
sunt coliniare în această ordine, astfel încât
AB=2 cm și BC=3 cm\def\arraystretch{1.5} \def,{{\char`,}} \def\Div{{\space\raisebox{-0.1em}{$\vdots$}\space}} {AB = 2 \text{ cm} \text{ și } BC = 3 \text{ cm}}
. Punctele
M și N\def\arraystretch{1.5} \def,{{\char`,}} \def\Div{{\space\raisebox{-0.1em}{$\vdots$}\space}} {M \text{ și } N}
sunt mijloacele segmentelor
AB\def\arraystretch{1.5} \def,{{\char`,}} \def\Div{{\space\raisebox{-0.1em}{$\vdots$}\space}} {AB}
, respectiv
AC\def\arraystretch{1.5} \def,{{\char`,}} \def\Div{{\space\raisebox{-0.1em}{$\vdots$}\space}} {AC}
. Lungimea segmentului
MN\def\arraystretch{1.5} \def,{{\char`,}} \def\Div{{\space\raisebox{-0.1em}{$\vdots$}\space}} {MN}
este egală cu:

A
0,5 cm\def\arraystretch{1.5} \def,{{\char`,}} \def\Div{{\space\raisebox{-0.1em}{$\vdots$}\space}} {0,5 \text{ cm}}
B
1 cm\def\arraystretch{1.5} \def,{{\char`,}} \def\Div{{\space\raisebox{-0.1em}{$\vdots$}\space}} {1 \text{ cm}}
C
1,5 cm\def\arraystretch{1.5} \def,{{\char`,}} \def\Div{{\space\raisebox{-0.1em}{$\vdots$}\space}} {1,5 \text{ cm}}
D
2 cm\def\arraystretch{1.5} \def,{{\char`,}} \def\Div{{\space\raisebox{-0.1em}{$\vdots$}\space}} {2 \text{ cm}}
2.
5

În figura alăturată, unghiurile
AOB, BOC, COD, DOE și EOF\def\arraystretch{1.5} \def,{{\char`,}} \def\Div{{\space\raisebox{-0.1em}{$\vdots$}\space}} {AOB,\text{ }BOC,\text{ }COD,\text{ }DOE \text{ și } EOF}
sunt congruente, astfel încât punctele
A, O și F\def\arraystretch{1.5} \def,{{\char`,}} \def\Div{{\space\raisebox{-0.1em}{$\vdots$}\space}} {A,\text{ }O \text{ și } F}
sunt coliniare. Măsura unghiului
BOD\def\arraystretch{1.5} \def,{{\char`,}} \def\Div{{\space\raisebox{-0.1em}{$\vdots$}\space}} {BOD}
este egală cu:

A
90∘\def\arraystretch{1.5} \def,{{\char`,}} \def\Div{{\space\raisebox{-0.1em}{$\vdots$}\space}} {90^\circ}
B
72∘\def\arraystretch{1.5} \def,{{\char`,}} \def\Div{{\space\raisebox{-0.1em}{$\vdots$}\space}} {72^\circ}
C
36∘\def\arraystretch{1.5} \def,{{\char`,}} \def\Div{{\space\raisebox{-0.1em}{$\vdots$}\space}} {36^\circ}
D
30∘\def\arraystretch{1.5} \def,{{\char`,}} \def\Div{{\space\raisebox{-0.1em}{$\vdots$}\space}} {30^\circ}
3.
5

Punctele
D, E și F\def\arraystretch{1.5} \def,{{\char`,}} \def\Div{{\space\raisebox{-0.1em}{$\vdots$}\space}} {D,\text{ }E \text{ și } F}
sunt mijloacele laturilor triunghiului
ABC\def\arraystretch{1.5} \def,{{\char`,}} \def\Div{{\space\raisebox{-0.1em}{$\vdots$}\space}} {ABC}
cu perimetrul egal cu
36 cm\def\arraystretch{1.5} \def,{{\char`,}} \def\Div{{\space\raisebox{-0.1em}{$\vdots$}\space}} {36 \text{ cm}}
. Perimetrul triunghiului
DEF\def\arraystretch{1.5} \def,{{\char`,}} \def\Div{{\space\raisebox{-0.1em}{$\vdots$}\space}} {DEF}
este egal cu:

A
12 cm\def\arraystretch{1.5} \def,{{\char`,}} \def\Div{{\space\raisebox{-0.1em}{$\vdots$}\space}} {12 \text{ cm}}
B
16 cm\def\arraystretch{1.5} \def,{{\char`,}} \def\Div{{\space\raisebox{-0.1em}{$\vdots$}\space}} {16 \text{ cm}}
C
24 cm\def\arraystretch{1.5} \def,{{\char`,}} \def\Div{{\space\raisebox{-0.1em}{$\vdots$}\space}} {24 \text{ cm}}
D
18 cm\def\arraystretch{1.5} \def,{{\char`,}} \def\Div{{\space\raisebox{-0.1em}{$\vdots$}\space}} {18 \text{ cm}}
4.
5

Trapezul
ABCD\def\arraystretch{1.5} \def,{{\char`,}} \def\Div{{\space\raisebox{-0.1em}{$\vdots$}\space}} {ABCD}
are baza mare
AB\def\arraystretch{1.5} \def,{{\char`,}} \def\Div{{\space\raisebox{-0.1em}{$\vdots$}\space}} {AB}
egală cu diagonala
AC\def\arraystretch{1.5} \def,{{\char`,}} \def\Div{{\space\raisebox{-0.1em}{$\vdots$}\space}} {AC}
și
AD=DC=CB\def\arraystretch{1.5} \def,{{\char`,}} \def\Div{{\space\raisebox{-0.1em}{$\vdots$}\space}} {AD=DC=CB}
. Măsura unghiului
ABC\def\arraystretch{1.5} \def,{{\char`,}} \def\Div{{\space\raisebox{-0.1em}{$\vdots$}\space}} {ABC}
este egală cu:

A
60∘\def\arraystretch{1.5} \def,{{\char`,}} \def\Div{{\space\raisebox{-0.1em}{$\vdots$}\space}} {60^\circ}
B
75∘\def\arraystretch{1.5} \def,{{\char`,}} \def\Div{{\space\raisebox{-0.1em}{$\vdots$}\space}} {75^\circ}
C
36∘\def\arraystretch{1.5} \def,{{\char`,}} \def\Div{{\space\raisebox{-0.1em}{$\vdots$}\space}} {36^\circ}
D
72∘\def\arraystretch{1.5} \def,{{\char`,}} \def\Div{{\space\raisebox{-0.1em}{$\vdots$}\space}} {72^\circ}
5.
5

În figura alăturată sunt reprezentate două cercuri concentrice de centru
O\def\arraystretch{1.5} \def,{{\char`,}} \def\Div{{\space\raisebox{-0.1em}{$\vdots$}\space}} {O}
și raze
OA\def\arraystretch{1.5} \def,{{\char`,}} \def\Div{{\space\raisebox{-0.1em}{$\vdots$}\space}} {OA}
și
OB\def\arraystretch{1.5} \def,{{\char`,}} \def\Div{{\space\raisebox{-0.1em}{$\vdots$}\space}} {OB}
. Coarda
BC\def\arraystretch{1.5} \def,{{\char`,}} \def\Div{{\space\raisebox{-0.1em}{$\vdots$}\space}} {BC}
este tangentă în punctul
A\def\arraystretch{1.5} \def,{{\char`,}} \def\Div{{\space\raisebox{-0.1em}{$\vdots$}\space}} {A}
la cercul mai mic. Știind că
OA=6 cm\def\arraystretch{1.5} \def,{{\char`,}} \def\Div{{\space\raisebox{-0.1em}{$\vdots$}\space}} {OA = 6 \text{ cm}}
și
OB=10 cm\def\arraystretch{1.5} \def,{{\char`,}} \def\Div{{\space\raisebox{-0.1em}{$\vdots$}\space}} {OB=10 \text{ cm}}
, lungimea segmentului
BC\def\arraystretch{1.5} \def,{{\char`,}} \def\Div{{\space\raisebox{-0.1em}{$\vdots$}\space}} {BC}
este egală cu:

A
434 cm\def\arraystretch{1.5} \def,{{\char`,}} \def\Div{{\space\raisebox{-0.1em}{$\vdots$}\space}} {4\sqrt{34} \text{ cm}}
B
83 cm\def\arraystretch{1.5} \def,{{\char`,}} \def\Div{{\space\raisebox{-0.1em}{$\vdots$}\space}} {8\sqrt{3} \text{ cm}}
C
16 cm\def\arraystretch{1.5} \def,{{\char`,}} \def\Div{{\space\raisebox{-0.1em}{$\vdots$}\space}} {16 \text{ cm}}
D
102 cm\def\arraystretch{1.5} \def,{{\char`,}} \def\Div{{\space\raisebox{-0.1em}{$\vdots$}\space}} {10\sqrt{2} \text{ cm}}
6.
5

Piramida patrulateră regulată
SABCD\def\arraystretch{1.5} \def,{{\char`,}} \def\Div{{\space\raisebox{-0.1em}{$\vdots$}\space}} {SABCD}
are toate muchiile congruente. Măsura unghiului dintre dreptele
SA\def\arraystretch{1.5} \def,{{\char`,}} \def\Div{{\space\raisebox{-0.1em}{$\vdots$}\space}} {SA}
și
BC\def\arraystretch{1.5} \def,{{\char`,}} \def\Div{{\space\raisebox{-0.1em}{$\vdots$}\space}} {BC}
este de:

A
30∘\def\arraystretch{1.5} \def,{{\char`,}} \def\Div{{\space\raisebox{-0.1em}{$\vdots$}\space}} {30^\circ}
B
60∘\def\arraystretch{1.5} \def,{{\char`,}} \def\Div{{\space\raisebox{-0.1em}{$\vdots$}\space}} {60^\circ}
C
90∘\def\arraystretch{1.5} \def,{{\char`,}} \def\Div{{\space\raisebox{-0.1em}{$\vdots$}\space}} {90^\circ}
D
120∘\def\arraystretch{1.5} \def,{{\char`,}} \def\Div{{\space\raisebox{-0.1em}{$\vdots$}\space}} {120^\circ}

Subiectul al III-lea

1.
5

Media aritmetică a două numere naturale este
25\def\arraystretch{1.5} \def,{{\char`,}} \def\Div{{\space\raisebox{-0.1em}{$\vdots$}\space}} {25}
. Împărțind un număr la celălalt, se obține câtul
11\def\arraystretch{1.5} \def,{{\char`,}} \def\Div{{\space\raisebox{-0.1em}{$\vdots$}\space}} {11}
și restul egal cu
2\def\arraystretch{1.5} \def,{{\char`,}} \def\Div{{\space\raisebox{-0.1em}{$\vdots$}\space}} {2}
.

a.
2
Poate fi numărul mai mare egal cu
45\def\arraystretch{1.5} \def,{{\char`,}} \def\Div{{\space\raisebox{-0.1em}{$\vdots$}\space}} {45}
? Justificați răspunsul.
b.
3
Aflați cele două numere.
2.
5

Fie numărul
a=235−5−15−(1,8)−1\def\arraystretch{1.5} \def,{{\char`,}} \def\Div{{\space\raisebox{-0.1em}{$\vdots$}\space}} {a = \frac{2}{3\sqrt{5}} - \frac{\sqrt{5}-1}{\sqrt{5}} - (\sqrt{1,8})^{-1}}
.

a.
2
Arătați că
a=−1\def\arraystretch{1.5} \def,{{\char`,}} \def\Div{{\space\raisebox{-0.1em}{$\vdots$}\space}} {a = -1}
.
b.
3
Dacă
b=23−5\def\arraystretch{1.5} \def,{{\char`,}} \def\Div{{\space\raisebox{-0.1em}{$\vdots$}\space}} {b = 2\sqrt{3} - 5}
, calculați
(6a−b+23)2023\def\arraystretch{1.5} \def,{{\char`,}} \def\Div{{\space\raisebox{-0.1em}{$\vdots$}\space}} {(6a - b + 2\sqrt{3})^{2023}}
.
3.
5

Fie
E(x)=(x−6x2−25+xx−5−2x+5):2x+4x2−25\def\arraystretch{1.5} \def,{{\char`,}} \def\Div{{\space\raisebox{-0.1em}{$\vdots$}\space}} {E(x)=\left( \frac{x-6}{x^2-25} + \frac{x}{x-5} - \frac{2}{x + 5} \right) : \frac{2x+4}{x^2-25}}
,
x∈R\{−5;−2;5}\def\arraystretch{1.5} \def,{{\char`,}} \def\Div{{\space\raisebox{-0.1em}{$\vdots$}\space}} {x \in \mathbb{R} \backslash \{-5; -2; 5\}}

a.
2
Arătați că
E(x)=x+22\def\arraystretch{1.5} \def,{{\char`,}} \def\Div{{\space\raisebox{-0.1em}{$\vdots$}\space}} {E(x)=\frac{x+2}{2}}
.
b.
3
Rezolvați în mulțimea numerelor întregi ecuația
∣4⋅E(x)−5∣=5\def\arraystretch{1.5} \def,{{\char`,}} \def\Div{{\space\raisebox{-0.1em}{$\vdots$}\space}} {|4 \cdot E(x) - 5| = 5}
.
4.
5

În exteriorul triunghiului dreptunghic
ABC\def\arraystretch{1.5} \def,{{\char`,}} \def\Div{{\space\raisebox{-0.1em}{$\vdots$}\space}} {ABC}
, cu catetele
AB=12 cm\def\arraystretch{1.5} \def,{{\char`,}} \def\Div{{\space\raisebox{-0.1em}{$\vdots$}\space}} {AB=12 \text{ cm}}
și
AC=16 cm\def\arraystretch{1.5} \def,{{\char`,}} \def\Div{{\space\raisebox{-0.1em}{$\vdots$}\space}} {AC=16 \text{ cm}}
, se construiesc triunghiurile dreptunghice isoscele
ABD\def\arraystretch{1.5} \def,{{\char`,}} \def\Div{{\space\raisebox{-0.1em}{$\vdots$}\space}} {ABD}
și
ACE\def\arraystretch{1.5} \def,{{\char`,}} \def\Div{{\space\raisebox{-0.1em}{$\vdots$}\space}} {ACE}
cu ipotenuzele
AB\def\arraystretch{1.5} \def,{{\char`,}} \def\Div{{\space\raisebox{-0.1em}{$\vdots$}\space}} {AB}
și respectiv
AC\def\arraystretch{1.5} \def,{{\char`,}} \def\Div{{\space\raisebox{-0.1em}{$\vdots$}\space}} {AC}
.

a.
2
Arătați că punctele
D, A și E\def\arraystretch{1.5} \def,{{\char`,}} \def\Div{{\space\raisebox{-0.1em}{$\vdots$}\space}} {D,\text{ }A \text{ și } E}
sunt puncte coliniare.
b.
3
Dacă
M\def\arraystretch{1.5} \def,{{\char`,}} \def\Div{{\space\raisebox{-0.1em}{$\vdots$}\space}} {M}
este mijlocul laturii
BC\def\arraystretch{1.5} \def,{{\char`,}} \def\Div{{\space\raisebox{-0.1em}{$\vdots$}\space}} {BC}
,
DM∩AB={P}\def\arraystretch{1.5} \def,{{\char`,}} \def\Div{{\space\raisebox{-0.1em}{$\vdots$}\space}} {DM \cap AB = \{P\}}
și
EM∩AC={Q}\def\arraystretch{1.5} \def,{{\char`,}} \def\Div{{\space\raisebox{-0.1em}{$\vdots$}\space}} {EM \cap AC = \{Q\}}
, calculați aria patrulaterului
APMQ\def\arraystretch{1.5} \def,{{\char`,}} \def\Div{{\space\raisebox{-0.1em}{$\vdots$}\space}} {APMQ}
.
5.
5

În figura alăturată sunt reprezentate două pătrate,
ABCD\def\arraystretch{1.5} \def,{{\char`,}} \def\Div{{\space\raisebox{-0.1em}{$\vdots$}\space}} {ABCD}
și
CDFE\def\arraystretch{1.5} \def,{{\char`,}} \def\Div{{\space\raisebox{-0.1em}{$\vdots$}\space}} {CDFE}
, având laturile de
18 cm\def\arraystretch{1.5} \def,{{\char`,}} \def\Div{{\space\raisebox{-0.1em}{$\vdots$}\space}} {18 \text{ cm}}
. Punctul
M\def\arraystretch{1.5} \def,{{\char`,}} \def\Div{{\space\raisebox{-0.1em}{$\vdots$}\space}} {M}
este mijlocul lui
CE\def\arraystretch{1.5} \def,{{\char`,}} \def\Div{{\space\raisebox{-0.1em}{$\vdots$}\space}} {CE}
, iar
AM∩CD={N}\def\arraystretch{1.5} \def,{{\char`,}} \def\Div{{\space\raisebox{-0.1em}{$\vdots$}\space}} {AM \cap CD = \{N\}}
.

a.
2
Arătați că
NC=6 cm\def\arraystretch{1.5} \def,{{\char`,}} \def\Div{{\space\raisebox{-0.1em}{$\vdots$}\space}} {NC = 6 \text{ cm}}
.
b.
3
Calculați sinusul unghiului
CAM\def\arraystretch{1.5} \def,{{\char`,}} \def\Div{{\space\raisebox{-0.1em}{$\vdots$}\space}} {CAM}
.
6.
5

În figura alăturată este reprezentat un paralelipiped dreptunghic
ABCDEFGH\def\arraystretch{1.5} \def,{{\char`,}} \def\Div{{\space\raisebox{-0.1em}{$\vdots$}\space}} {ABCDEFGH}
cu
AB=20 cm\def\arraystretch{1.5} \def,{{\char`,}} \def\Div{{\space\raisebox{-0.1em}{$\vdots$}\space}} {AB = 20 \text{ cm}}
,
BC=15 cm\def\arraystretch{1.5} \def,{{\char`,}} \def\Div{{\space\raisebox{-0.1em}{$\vdots$}\space}} {BC = 15 \text{ cm}}
și
ACGE\def\arraystretch{1.5} \def,{{\char`,}} \def\Div{{\space\raisebox{-0.1em}{$\vdots$}\space}} {ACGE}
este pătrat.

a.
2
Arătați că
AE=25 cm\def\arraystretch{1.5} \def,{{\char`,}} \def\Div{{\space\raisebox{-0.1em}{$\vdots$}\space}} {AE = 25 \text{ cm}}
.
b.
3
Punctele
M\def\arraystretch{1.5} \def,{{\char`,}} \def\Div{{\space\raisebox{-0.1em}{$\vdots$}\space}} {M}
și
N\def\arraystretch{1.5} \def,{{\char`,}} \def\Div{{\space\raisebox{-0.1em}{$\vdots$}\space}} {N}
aparțin segmentelor
BF\def\arraystretch{1.5} \def,{{\char`,}} \def\Div{{\space\raisebox{-0.1em}{$\vdots$}\space}} {BF}
și respectiv
EG\def\arraystretch{1.5} \def,{{\char`,}} \def\Div{{\space\raisebox{-0.1em}{$\vdots$}\space}} {EG}
, astfel încât
BM=EN=5 cm\def\arraystretch{1.5} \def,{{\char`,}} \def\Div{{\space\raisebox{-0.1em}{$\vdots$}\space}} {BM = EN = 5 \text{ cm}}
. Calculați tangenta unghiului format de dreapta
MN\def\arraystretch{1.5} \def,{{\char`,}} \def\Div{{\space\raisebox{-0.1em}{$\vdots$}\space}} {MN}
cu planul
(BFC)\def\arraystretch{1.5} \def,{{\char`,}} \def\Div{{\space\raisebox{-0.1em}{$\vdots$}\space}} {(BFC)}
.
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