Prvo rješenje.
Povucimo visinu iz vrha C C C A B AB A B N N N ∠ B A C \angle BAC ∠ B A C ∠ A B C \angle ABC ∠ A B C C A N CAN C A N C B N CBN C B N
1 = tg ∠ B A C = tg ∠ N A C = ∣ C N ∣ ∣ A N ∣ , 1=\operatorname{tg}\angle BAC=\operatorname{tg}\angle NAC=\frac{|CN|}{|AN|}, 1 = tg ∠ B A C = tg ∠ N A C = ∣ A N ∣ ∣ C N ∣ , te
2 = tg ∠ A B C = tg ∠ N B C = ∣ C N ∣ ∣ B N ∣ . 2=\operatorname{tg}\angle ABC=\operatorname{tg}\angle NBC=\frac{|CN|}{|BN|}. 2 = tg ∠ A B C = tg ∠ N B C = ∣ B N ∣ ∣ C N ∣ . Slijedi ∣ B N ∣ : ∣ A N ∣ = 1 : 2 |BN|:|AN|=1:2 ∣ B N ∣ : ∣ A N ∣ = 1 : 2
Označimo ∣ B N ∣ = x |BN|=x ∣ B N ∣ = x ∣ A N ∣ = ∣ C N ∣ = 2 x |AN|=|CN|=2x ∣ A N ∣ = ∣ C N ∣ = 2 x
∣ A B ∣ = ∣ A N ∣ + ∣ N B ∣ = 2 x + x = 3 x , |AB|=|AN|+|NB|=2x+x=3x, ∣ A B ∣ = ∣ A N ∣ + ∣ N B ∣ = 2 x + x = 3 x , a iz Pitagorinog poučka primijenjenog na pravokutni trokut C B N CBN C B N
∣ B C ∣ = ∣ C N ∣ 2 + ∣ B N ∣ 2 = 4 x 2 + x 2 = 5 x . |BC|=\sqrt{|CN|^2+|BN|^2}=\sqrt{4x^2+x^2}=\sqrt5x. ∣ B C ∣ = ∣ C N ∣ 2 + ∣ B N ∣ 2 = 4 x 2 + x 2 = 5 x . Zaključujemo da je
∣ B C ∣ : ∣ A B ∣ = 5 x : 3 x = 5 : 3. |BC|:|AB|=\sqrt5x:3x=\sqrt5:3. ∣ B C ∣ : ∣ A B ∣ = 5 x : 3 x = 5 : 3. Drugo rješenje.
Neka je a = ∣ B C ∣ a=|BC| a = ∣ B C ∣ b = ∣ A C ∣ b=|AC| b = ∣ A C ∣ c = ∣ A B ∣ c=|AB| c = ∣ A B ∣ ∠ C A B = α \angle CAB=\alpha ∠ C A B = α ∠ A B C = β \angle ABC=\beta ∠ A B C = β ∠ B C A = γ \angle BCA=\gamma ∠ B C A = γ
Izračunajmo tg γ \operatorname{tg}\gamma tg γ
tg ( α + β ) = tg α + tg β 1 − tg α tg β = 1 + 2 1 − 1 ⋅ 2 = − 3. \operatorname{tg}(\alpha+\beta)=\frac{\operatorname{tg}\alpha+\operatorname{tg}\beta}{1-\operatorname{tg}\alpha\operatorname{tg}\beta}=\frac{1+2}{1-1\cdot2}=-3. tg ( α + β ) = 1 − tg α tg β tg α + tg β = 1 − 1 ⋅ 2 1 + 2 = − 3. Slijedi da je tg γ = tg ( 180 ∘ − α − β ) = − tg ( α + β ) = 3 \operatorname{tg}\gamma=\operatorname{tg}(180^\circ-\alpha-\beta)=-\operatorname{tg}(\alpha+\beta)=3 tg γ = tg ( 18 0 ∘ − α − β ) = − tg ( α + β ) = 3
Prema poučku o sinusima vrijedi
a c = sin α sin γ . \frac{a}{c}=\frac{\sin\alpha}{\sin\gamma}. c a = sin γ sin α . Koristeći poučak o sinusima i poučak o kosinusu dobivamo
tg α tg β = sin α sin β ⋅ cos β cos α = a b ⋅ a 2 + c 2 − b 2 2 a c ⋅ 2 b c b 2 + c 2 − a 2 = a 2 + c 2 − b 2 b 2 + c 2 − a 2 . \frac{\operatorname{tg}\alpha}{\operatorname{tg}\beta}=\frac{\sin\alpha}{\sin\beta}\cdot\frac{\cos\beta}{\cos\alpha}=\frac{a}{b}\cdot\frac{a^2+c^2-b^2}{2ac}\cdot\frac{2bc}{b^2+c^2-a^2}=\frac{a^2+c^2-b^2}{b^2+c^2-a^2}. tg β tg α = sin β sin α ⋅ cos α cos β = b a ⋅ 2 a c a 2 + c 2 − b 2 ⋅ b 2 + c 2 − a 2 2 b c = b 2 + c 2 − a 2 a 2 + c 2 − b 2 . Budući da je tg α = 1 \operatorname{tg}\alpha=1 tg α = 1 tg β = 2 \operatorname{tg}\beta=2 tg β = 2
b 2 + c 2 − a 2 = 2 ( a 2 + c 2 − b 2 ) , b^2+c^2-a^2=2(a^2+c^2-b^2), b 2 + c 2 − a 2 = 2 ( a 2 + c 2 − b 2 ) , tj. 3 b 2 = c 2 + 3 a 2 3b^2=c^2+3a^2 3 b 2 = c 2 + 3 a 2
Drugu relaciju između a a a b b b c c c
tg α tg γ = b 2 + a 2 − c 2 b 2 + c 2 − a 2 . \frac{\operatorname{tg}\alpha}{\operatorname{tg}\gamma}=\frac{b^2+a^2-c^2}{b^2+c^2-a^2}. tg γ tg α = b 2 + c 2 − a 2 b 2 + a 2 − c 2 . Uvrstimo li tg α = 1 \operatorname{tg}\alpha=1 tg α = 1 tg γ = 3 \operatorname{tg}\gamma=3 tg γ = 3
b 2 + c 2 − a 2 = 3 ( b 2 + a 2 − c 2 ) , b^2+c^2-a^2=3(b^2+a^2-c^2), b 2 + c 2 − a 2 = 3 ( b 2 + a 2 − c 2 ) , tj. 2 c 2 = 2 a 2 + b 2 2c^2=2a^2+b^2 2 c 2 = 2 a 2 + b 2
Eliminacijom varijable b b b 3 b 2 = c 2 + 3 a 2 3b^2=c^2+3a^2 3 b 2 = c 2 + 3 a 2 2 c 2 = 2 a 2 + b 2 2c^2=2a^2+b^2 2 c 2 = 2 a 2 + b 2 9 a 2 = 5 c 2 9a^2=5c^2 9 a 2 = 5 c 2 a : c = 5 : 3 a:c=\sqrt5:3 a : c = 5 : 3
Treće rješenje.
Neka je a = ∣ B C ∣ a=|BC| a = ∣ B C ∣ b = ∣ A C ∣ b=|AC| b = ∣ A C ∣ c = ∣ A B ∣ c=|AB| c = ∣ A B ∣ ∠ C A B = α \angle CAB=\alpha ∠ C A B = α ∠ A B C = β \angle ABC=\beta ∠ A B C = β ∠ B C A = γ \angle BCA=\gamma ∠ B C A = γ tg γ = 3 \operatorname{tg}\gamma=3 tg γ = 3
Prema poučku o sinusima vrijedi
a c = sin α sin γ . \frac{a}{c}=\frac{\sin\alpha}{\sin\gamma}. c a = sin γ sin α . Želimo izraziti sin α \sin\alpha sin α sin γ \sin\gamma sin γ tg α \operatorname{tg}\alpha tg α tg γ \operatorname{tg}\gamma tg γ
1 = tg α = sin α cos α , 1=\operatorname{tg}\alpha=\frac{\sin\alpha}{\cos\alpha}, 1 = tg α = cos α sin α , nakon kvadriranja i uvrštavanja cos 2 α = 1 − sin 2 α \cos^2\alpha=1-\sin^2\alpha cos 2 α = 1 − sin 2 α 1 − sin 2 α = sin 2 α 1-\sin^2\alpha=\sin^2\alpha 1 − sin 2 α = sin 2 α 2 sin 2 α = 1 2\sin^2\alpha=1 2 sin 2 α = 1
Analogno, budući da je 3 = tg γ 3=\operatorname{tg}\gamma 3 = tg γ 9 ( 1 − sin 2 γ ) = sin 2 γ 9(1-\sin^2\gamma)=\sin^2\gamma 9 ( 1 − sin 2 γ ) = sin 2 γ 10 sin 2 γ = 9 10\sin^2\gamma=9 10 sin 2 γ = 9
Slijedi da je
a 2 c 2 = sin 2 α sin 2 γ = 5 9 , \frac{a^2}{c^2}=\frac{\sin^2\alpha}{\sin^2\gamma}=\frac{5}{9}, c 2 a 2 = sin 2 γ sin 2 α = 9 5 , tj. a : c = 5 : 3 a:c=\sqrt5:3 a : c = 5 : 3
Četvrto rješenje.
Neka je a = ∣ B C ∣ a=|BC| a = ∣ B C ∣ b = ∣ A C ∣ b=|AC| b = ∣ A C ∣ c = ∣ A B ∣ c=|AB| c = ∣ A B ∣ ∠ C A B = α \angle CAB=\alpha ∠ C A B = α ∠ A B C = β \angle ABC=\beta ∠ A B C = β ∠ B C A = γ \angle BCA=\gamma ∠ B C A = γ
Prema poučku o sinusima vrijedi
a c = sin α sin γ . \frac{a}{c}=\frac{\sin\alpha}{\sin\gamma}. c a = sin γ sin α . Vrijedi sin γ = sin ( α + β ) = sin α cos β + cos α sin β \sin\gamma=\sin(\alpha+\beta)=\sin\alpha\cos\beta+\cos\alpha\sin\beta sin γ = sin ( α + β ) = sin α cos β + cos α sin β
sin γ sin α = cos β + sin β tg α = sin β ( 1 tg β + 1 tg α ) . \frac{\sin\gamma}{\sin\alpha}=\cos\beta+\frac{\sin\beta}{\operatorname{tg}\alpha}=\sin\beta\left(\frac{1}{\operatorname{tg}\beta}+\frac{1}{\operatorname{tg}\alpha}\right). sin α sin γ = cos β + tg α sin β = sin β ( tg β 1 + tg α 1 ) . Budući da je 2 = tg β = sin β cos β 2=\operatorname{tg}\beta=\frac{\sin\beta}{\cos\beta} 2 = tg β = c o s β s i n β cos 2 β = 1 − sin 2 β \cos^2\beta=1-\sin^2\beta cos 2 β = 1 − sin 2 β 4 ( 1 − sin 2 β ) = sin 2 β 4(1-\sin^2\beta)=\sin^2\beta 4 ( 1 − sin 2 β ) = sin 2 β 5 sin 2 β = 4 5\sin^2\beta=4 5 sin 2 β = 4
Budući da je β \beta β sin β = 2 5 \sin\beta=\frac{2}{\sqrt5} sin β = 5 2
Slijedi da je
sin γ sin α = 2 5 ( 1 2 + 1 ) = 3 5 \frac{\sin\gamma}{\sin\alpha}=\frac{2}{\sqrt5}\left(\frac12+1\right)=\frac{3}{\sqrt5} sin α sin γ = 5 2 ( 2 1 + 1 ) = 5 3 i konačno, traženi omjer je 5 3 \frac{\sqrt5}{3} 3 5
