+583 In the following fighure ,if triangle ABC is a right triangle,angle A= 90 degrees,point D located on AC ,point E is loacted on BC, $${\mathtt{AB}} = {\frac{{\mathtt{1}}}{{\mathtt{2}}}}{\mathtt{\,\times\,}}{\mathtt{BD}}$$ ,and $${\mathtt{CE}} = {\frac{{\mathtt{1}}}{{\mathtt{4}}}}{\mathtt{\,\times\,}}{\mathtt{EB}}$$. angle BDE=120 degrees,CD=3, then BC=?
+26367 In the following fighure ,if triangle ABC is a right triangle,angle A= 90 degrees,point D located on AC ,point E is loacted on BC, $${\mathtt{AB}} = {\frac{{\mathtt{1}}}{{\mathtt{2}}}}{\mathtt{\,\times\,}}{\mathtt{BD}}$$ ,and $${\mathtt{CE}} = {\frac{{\mathtt{1}}}{{\mathtt{4}}}}{\mathtt{\,\times\,}}{\mathtt{EB}}$$. angle BDE=120 degrees,CD=3, then BC=?
We define: CD = 3, AB = y, DB = 2y, CE = x, EB = 4x, BC = 5x, ED = u, DA = d
1. Pythagoras:
$$\small{\overline{AB}^2 + \overline{DA}^2 = \overline{DB}^2}\\
\small{y^2+d^2=4y^2\qquad d^2 = 3y^2 \qquad d = \sqrt{3}y }$$
2. Pythagoras:
$$\small{\overline{CA}^2 + \overline{AB}^2 = \overline{BC}^2 \qquad \overline{CA} = 3+\sqrt{3}y}\\\\
\small{ (3+\sqrt{3}y)^2+y^2=(5x)^2\qquad \cdots \qquad
\boxed{x = \dfrac{ \sqrt{ 4y^2+6\sqrt{3}y+9 } } {5} ~~(1) }}$$
3. cos-rule:
$$\small{u^2 = 3^2 + x^2 - 2\cdot 3\cdot x \cdot \cos{(C)}\qquad \cos{(C)} = \dfrac{ 3 + \sqrt{3}y }{5x} \quad \cdots \quad \boxed{ u = \sqrt{x^2-1.2\sqrt{3}+5.4} ~~(2)}}$$
4. cos-rule:
$$\small{
\begin{array}{lcl}
(4x)^2 = u^2+(2y)^2-2u2y\cos{ (120\ensurement{^{\circ}}) } \quad \cos{(120\ensurement{^{\circ}}) } = -\frac{1}{2} \quad \cdots \quad 16x^2=u^2+4y^2+2uy \\
\text{we substitute u, formula (2)}\\
\cdots\\
15x^2 = -1.2\sqrt{3}y+5.4+4y^2+2y\sqrt{x^2-1.2\sqrt{3}y+5.4}\\
\text{we substitute first x, formula (1)}\\
\cdots\\
\frac{12}{5}y^2+}\frac{18}{5}\sqrt{3}y=4y^2-1.2\sqrt{3}y+2y\sqrt{x^2-1.2\sqrt{3}y+5.4} \qaud | \quad :y \\
\frac{12}{5}y+}\frac{18}{5}\sqrt{3}=4y-1.2\sqrt{3}+2\sqrt{x^2-1.2\sqrt{3}y+5.4} \\
\cdots\\
x^2-1.2\sqrt{3}y+5.4=(2.4\sqrt{3}-0.8y)^2\\
\text{we substitute again x, formula (1)}\\
\cdots\\
0.16y^2+0.24\sqrt{3}y+0.36-1.2\sqrt{3}y+5.4=17.28-3.84\sqrt{3}y+0.64y^2\\
0.48y^2-2.88\sqrt{3}y+11.52=0 \quad | :0.48 \quad \\
\boxed{y^2-6\sqrt{3}+24 = 0} \\\\
y_{1,2} = \frac{6\sqrt{3}\pm \2\sqrt{3}}{2}\\
y_1 = 4\sqrt{3} \quad \text{no solution}\\
y_2 = 2\sqrt{3} \quad \text{solution}\\\\
x = \dfrac{ \sqrt{ 4y^2+6\sqrt{3}y+9 } } {5} \\\\
x = \dfrac{ \sqrt{ 4 (2\sqrt{3})^2+6\sqrt{3}( 2\sqrt{3})+9 } } {5} \\\\
x = \dfrac{ \sqrt{ 48 +36+9 } } {5} \\\\
x = 1.92873015220\\\\
\mathbf{ \overline{BC} =5x = 9.64365076099 }
\end{array}
}$$
+26367 In the following fighure ,if triangle ABC is a right triangle,angle A= 90 degrees,point D located on AC ,point E is loacted on BC, $${\mathtt{AB}} = {\frac{{\mathtt{1}}}{{\mathtt{2}}}}{\mathtt{\,\times\,}}{\mathtt{BD}}$$ ,and $${\mathtt{CE}} = {\frac{{\mathtt{1}}}{{\mathtt{4}}}}{\mathtt{\,\times\,}}{\mathtt{EB}}$$. angle BDE=120 degrees,CD=3, then BC=?
We define: CD = 3, AB = y, DB = 2y, CE = x, EB = 4x, BC = 5x, ED = u, DA = d
1. Pythagoras:
$$\small{\overline{AB}^2 + \overline{DA}^2 = \overline{DB}^2}\\
\small{y^2+d^2=4y^2\qquad d^2 = 3y^2 \qquad d = \sqrt{3}y }$$
2. Pythagoras:
$$\small{\overline{CA}^2 + \overline{AB}^2 = \overline{BC}^2 \qquad \overline{CA} = 3+\sqrt{3}y}\\\\
\small{ (3+\sqrt{3}y)^2+y^2=(5x)^2\qquad \cdots \qquad
\boxed{x = \dfrac{ \sqrt{ 4y^2+6\sqrt{3}y+9 } } {5} ~~(1) }}$$
3. cos-rule:
$$\small{u^2 = 3^2 + x^2 - 2\cdot 3\cdot x \cdot \cos{(C)}\qquad \cos{(C)} = \dfrac{ 3 + \sqrt{3}y }{5x} \quad \cdots \quad \boxed{ u = \sqrt{x^2-1.2\sqrt{3}+5.4} ~~(2)}}$$
4. cos-rule:
$$\small{
\begin{array}{lcl}
(4x)^2 = u^2+(2y)^2-2u2y\cos{ (120\ensurement{^{\circ}}) } \quad \cos{(120\ensurement{^{\circ}}) } = -\frac{1}{2} \quad \cdots \quad 16x^2=u^2+4y^2+2uy \\
\text{we substitute u, formula (2)}\\
\cdots\\
15x^2 = -1.2\sqrt{3}y+5.4+4y^2+2y\sqrt{x^2-1.2\sqrt{3}y+5.4}\\
\text{we substitute first x, formula (1)}\\
\cdots\\
\frac{12}{5}y^2+}\frac{18}{5}\sqrt{3}y=4y^2-1.2\sqrt{3}y+2y\sqrt{x^2-1.2\sqrt{3}y+5.4} \qaud | \quad :y \\
\frac{12}{5}y+}\frac{18}{5}\sqrt{3}=4y-1.2\sqrt{3}+2\sqrt{x^2-1.2\sqrt{3}y+5.4} \\
\cdots\\
x^2-1.2\sqrt{3}y+5.4=(2.4\sqrt{3}-0.8y)^2\\
\text{we substitute again x, formula (1)}\\
\cdots\\
0.16y^2+0.24\sqrt{3}y+0.36-1.2\sqrt{3}y+5.4=17.28-3.84\sqrt{3}y+0.64y^2\\
0.48y^2-2.88\sqrt{3}y+11.52=0 \quad | :0.48 \quad \\
\boxed{y^2-6\sqrt{3}+24 = 0} \\\\
y_{1,2} = \frac{6\sqrt{3}\pm \2\sqrt{3}}{2}\\
y_1 = 4\sqrt{3} \quad \text{no solution}\\
y_2 = 2\sqrt{3} \quad \text{solution}\\\\
x = \dfrac{ \sqrt{ 4y^2+6\sqrt{3}y+9 } } {5} \\\\
x = \dfrac{ \sqrt{ 4 (2\sqrt{3})^2+6\sqrt{3}( 2\sqrt{3})+9 } } {5} \\\\
x = \dfrac{ \sqrt{ 48 +36+9 } } {5} \\\\
x = 1.92873015220\\\\
\mathbf{ \overline{BC} =5x = 9.64365076099 }
\end{array}
}$$
