4- le triangler cer est il droit ? cos(frc)= rf/er = 9√3/15√3= 3/5 ⇒ cos(frc) = 0.6 ⇒ fcr = 53.16 degre cos(fce)=cf/ec = 5√3/13√3 = 5/13 ⇒fcr=67.38 degre
on 53+67 ≥≥90 don triangle n'est pas droit
exe 2 1- la longeur mt sin 18 = ct/mt ⇒ mt = ct/sin18 = 556/sin18 = 1799.25 m ⇒ mt≈≈1800 m 2- altitude du pic midi Ap Ap=Am+Ab+bp on calcul bp : sin12.1=bp/pt ⇒ bp= pt*sin12.1 = 494.699 ≈≈495 m
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a - on applique phytagore
1- er² = rf²+ef² = (9√3)²+(12√3)² = 81×3+144×3 = 3(81+144) = 3×225
⇒ er = √3×225 = √225√3 = 15√3
ec²=fc²+ef² = (5√3)²+(12√3)² = 25×3 + 144×3 = 3 ×169
⇒ec= √3×169 = √169√3 = 13√3
2 - perimetre de cer = p
p = er+ec+rc = er +ec +rf+fc = 15√3+13√3+9√3+5√3 = 42√3 = 72.74
p = 72.74
3- Air triangle = a
a=1/2 rc*ef = 1/2 rc ( rf+fc ) = 1/2 * 12√3 * ( 9√3+5√3) = 252
4- le triangler cer est il droit ?
cos(frc)= rf/er = 9√3/15√3= 3/5 ⇒ cos(frc) = 0.6 ⇒ fcr = 53.16 degre
cos(fce)=cf/ec = 5√3/13√3 = 5/13 ⇒fcr=67.38 degre
on 53+67 ≥≥90 don triangle n'est pas droit
exe 2
1- la longeur mt
sin 18 = ct/mt ⇒ mt = ct/sin18 = 556/sin18 = 1799.25 m
⇒ mt≈≈1800 m
2- altitude du pic midi Ap
Ap=Am+Ab+bp
on calcul bp : sin12.1=bp/pt ⇒ bp= pt*sin12.1 = 494.699 ≈≈495 m
Ap= 1753+556+495 = 2804 m
ex3 :
a=√20-12√5+2√125 = √4×5 - 12√5 + 2√25×5 = 2√5 - 12√5 +10√5 = 0
b=5√27-3√3+√12 = 5√9×3 - 3√3 + √4×3 = 15√3-3√3+2√3 = 14√3
c=√12 +2√48 -√75 = √4×3 + 2√3×16 -√25×3 = 5√3
d = √45 -2√5+√500 = √9×5 -2√5+√5×100 = 11√5
e=√20 -√45 -7√5 = √4×5 -√9×5 -7√5 = -8√5
f=√27-3√75 = √9×3 - 3√3×25 = -12√5
g = √180 +3√80 -2√125 = √5×36 + 3√5×16 -2√25×5 = 8√5
ex4 :
a=(3√2-5)² = (3√2)²-2×3×√2 ×5 +5³ = 43 -30√2
b=(4√7+2)² = (4√7)² +2×4√7×2 +2² = 116+16√7
c=(2√6-1)(2√6+1) = (2√6)²-1² = 23
d= (8-2√3)²= 8² -2×8×2√3 +(2√3)² = 76 - 32√3
e=(4+3√6)² = 4²+2×4×3√6 +(3√6)² = 70 +24√6
f=(5√2-2√3)(5√2+2√3)= (5√2)²-(2√3)² = 50-12 = 48