Question

Da figura sabemos que:

* B^CD é reto;
* BD é bissetriz de A^BC;
* BC mede 6 cm;

Determine a medida de AB.

Solução: 2 . ( 6 - 2√3 )cm

Favor apresentar os cálculos

Answer 1

Triângulo ABD

Soma dos ângulos internos de um triângulo é 180 graus. Logo:

105 graus + 45 graus + B = 180 graus

B + 150 = 180

b = 180 – 150

B = 30 graus

 

Triângulo retângulo BCD

Como BD é bissetriz do ângulo B temos:

B = 30 graus

BC = 6 cm

 

Aplicando o cosseno do ângulo B (30 graus) teremos:

cos 30 graus = BC/BD

V3/2 = 6/BD

BD = 6 : V3/2

BD = 4V3 cm

 

No triângulo ABD temos os seguintes elementos:

A = 105 graus

B = 30 graus

D = 45 graus

BD = 4V3 cm

 

Aplicando a Leio dos Senos encontraremos:

BD/sen 105 = AB/sen 45

Onde:

BD = 4V3 cm

AB = ?

sen 105 = sen (60 + 45) = sen 60 . cos 45 + sen 45 . cos 60 = (V6 + V2)/4

sen 45 = V2/2

4V3 : (V6 + V2)/4 = AB : V2/2

AB = (4V3 . V2/2) : (V6 + V2)/4

AB = (2V6) . (4/V6 + V2)

AB = (8V6)/(V6 + V2)

AB = [(8V6) . (V6 – V2)]/[(V6 + V2) . (V6 – V2)]

AB = [(8V36 – 8V12)]/6 – 2

AB = [(8.6 – 8.2V3)]/4

AB = [(48 – 8.2V3)]/4

AB = 12 – 4V3

AB = 2 . (6 – 2V3) cm

 

Resposta: 2 . (6 – 2V3) cm.

Bons estudos!

Answer 2

A soma dos ângulos internos de um triângulo é $180^{\circ}$. Assim, no triângulo $\text{ABD}$, temos:

$\measuredangle\text{ABD}+105^{\circ}+45^{\circ}=180^{\circ} \iff \measuredangle\text{ABD}+150^{\circ}=180^{\circ}$

$\measuredangle\text{ABD}=180^{\circ}-150^{\circ} \iff \measuredangle\text{ABD}=30^{\circ}$

Como $\text{BC}$ é bissetriz de $\measuredangle\text{ABC}$, então $\measuredangle \text{CBD}=\measuredangle\text{ABD}$

Deste modo, no triângulo $\text {BCD}$, obtemos:

$\text{cos}~30^{\circ}=\dfrac{\text{BC}}{\text{BD}} \iff \dfrac{\sqrt{3}}{2}=\dfrac{6}{\text{BD}}$

$\sqrt{3}\cdot\text{BD}=12 \iff \text{BD}=\dfrac{12}{\sqrt{3}}$

$\text{BD}=\dfrac{12}{\sqrt{3}}\cdot\dfrac{\sqrt{3}}{\sqrt{3}} \iff \text{BF}=\dfrac{12\sqrt{3}}{3}$

Logo, $\text{BD}=4\sqrt{3}~\text{cm}$

Pela Lei dos Senos, no triângulo $\text{ABD}$:

$\dfrac{\text{AB}}{\text{sen}~45^{\circ}}=\dfrac{\text{BD}}{\text{sen}~105^{\circ}}$

Antes precisamos determinar $\text{sen}~105^{\circ}$. Para isso, vamos utilizar soma de arcos $\tex{sen}~(a+b)=\text{sen}~a\cdot\text{cos}~b+\text{sen}~b\cdot\text{cos}~a$

$\text{sen}~105^{\circ}=\text{sen}~45^{\circ}\cdot\text{cos}~60^{\circ}+\text{sen}~60^{\circ}\cdot\text{cos}~45^{\circ}$

$\text{sen}~105^{\circ}=\dfrac{\sqrt{2}}{2}\cdot\dfrac{1}{2}+\dfrac{\sqrt{3}}{2}\cdot\dfrac{\sqrt{2}}{2}$

$\text{sen}~105^{\circ}=\dfrac{\sqrt{2}}{4}+\dfrac{\sqrt{6}}{4} \iff \text{sen}~105^{\circ}=\dfrac{\sqrt{6}+\sqrt{2}}{4}$

Logo:

$\dfrac{\text{AB}}{\frac{\sqrt{2}}{2}}=\dfrac{4\sqrt{3}}{\frac{\sqrt{6}+\sqrt{2}}{4}} \iff \text{AB}\cdot\dfrac{\sqrt{6}+\sqrt{2}}{4}=\dfrac{4\sqrt{6}}{2}$

$\text{AB}=\dfrac{\dfrac{4\sqrt{6}}{2}}{\dfrac{\sqrt{6}+\sqrt{2}}{4}} \iff \text{AB}=\dfrac{4\sqrt{6}}{2}\cdot\dfrac{4}{\sqrt{6}+\sqrt{2}}$

$\text{AB}=\dfrac{8\sqrt{6}}{\sqrt{6}+\sqrt{2}}$

Racionalizando:

$\text{AB}=\dfrac{8\sqrt{6}}{\sqrt{6}+\sqrt{2}}\cdot\dfrac{\sqrt{6}-\sqrt{2}}{\sqrt{6}-\sqrt{2}}=\dfrac{8\cdot6 - 8\sqrt{12}}{(\sqrt{6})^2-(\sqrt{2})^2}$

$\text{AB}=\dfrac{48 - 8\cdot2\sqrt{3}}{6-2} \iff \text{AB}=\dfrac{48-16\sqrt{3}}{4}$

$\text{AB}=12 - 4\sqrt{3}$

Colocando $2$ em evidência:

$\text{AB}=2\cdot(6-2\sqrt{3})~\text{cm}$