+0

triangle

+8
2
1550
6
+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=?

Jun 24, 2015

#1
+26319
+18

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:

$$\small{ \begin{array}{lcl} (4x)^2 = u^2+(2y)^2-2u2y\cos{ (120\ensurement{^{\circ}}) } \quad \cos{(120\ensurement{^{\circ}}) } = -\frac{1}{2} \quad \cdots \quad 16x^2=u^2+4y^2+2uy \\ \text{we substitute u, formula (2)}\\ \cdots\\ 15x^2 = -1.2\sqrt{3}y+5.4+4y^2+2y\sqrt{x^2-1.2\sqrt{3}y+5.4}\\ \text{we substitute first x, formula (1)}\\ \cdots\\ \frac{12}{5}y^2+}\frac{18}{5}\sqrt{3}y=4y^2-1.2\sqrt{3}y+2y\sqrt{x^2-1.2\sqrt{3}y+5.4} \qaud | \quad :y \\ \frac{12}{5}y+}\frac{18}{5}\sqrt{3}=4y-1.2\sqrt{3}+2\sqrt{x^2-1.2\sqrt{3}y+5.4} \\ \cdots\\ x^2-1.2\sqrt{3}y+5.4=(2.4\sqrt{3}-0.8y)^2\\ \text{we substitute again x, formula (1)}\\ \cdots\\ 0.16y^2+0.24\sqrt{3}y+0.36-1.2\sqrt{3}y+5.4=17.28-3.84\sqrt{3}y+0.64y^2\\ 0.48y^2-2.88\sqrt{3}y+11.52=0 \quad | :0.48 \quad \\ \boxed{y^2-6\sqrt{3}+24 = 0} \\\\ y_{1,2} = \frac{6\sqrt{3}\pm \2\sqrt{3}}{2}\\ y_1 = 4\sqrt{3} \quad \text{no solution}\\ y_2 = 2\sqrt{3} \quad \text{solution}\\\\ x = \dfrac{ \sqrt{ 4y^2+6\sqrt{3}y+9 } } {5} \\\\ x = \dfrac{ \sqrt{ 4 (2\sqrt{3})^2+6\sqrt{3}( 2\sqrt{3})+9 } } {5} \\\\ x = \dfrac{ \sqrt{ 48 +36+9 } } {5} \\\\ x = 1.92873015220\\\\ \mathbf{ \overline{BC} =5x = 9.64365076099 } \end{array} }$$

Jun 24, 2015

#1
+26319
+18

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:

$$\small{ \begin{array}{lcl} (4x)^2 = u^2+(2y)^2-2u2y\cos{ (120\ensurement{^{\circ}}) } \quad \cos{(120\ensurement{^{\circ}}) } = -\frac{1}{2} \quad \cdots \quad 16x^2=u^2+4y^2+2uy \\ \text{we substitute u, formula (2)}\\ \cdots\\ 15x^2 = -1.2\sqrt{3}y+5.4+4y^2+2y\sqrt{x^2-1.2\sqrt{3}y+5.4}\\ \text{we substitute first x, formula (1)}\\ \cdots\\ \frac{12}{5}y^2+}\frac{18}{5}\sqrt{3}y=4y^2-1.2\sqrt{3}y+2y\sqrt{x^2-1.2\sqrt{3}y+5.4} \qaud | \quad :y \\ \frac{12}{5}y+}\frac{18}{5}\sqrt{3}=4y-1.2\sqrt{3}+2\sqrt{x^2-1.2\sqrt{3}y+5.4} \\ \cdots\\ x^2-1.2\sqrt{3}y+5.4=(2.4\sqrt{3}-0.8y)^2\\ \text{we substitute again x, formula (1)}\\ \cdots\\ 0.16y^2+0.24\sqrt{3}y+0.36-1.2\sqrt{3}y+5.4=17.28-3.84\sqrt{3}y+0.64y^2\\ 0.48y^2-2.88\sqrt{3}y+11.52=0 \quad | :0.48 \quad \\ \boxed{y^2-6\sqrt{3}+24 = 0} \\\\ y_{1,2} = \frac{6\sqrt{3}\pm \2\sqrt{3}}{2}\\ y_1 = 4\sqrt{3} \quad \text{no solution}\\ y_2 = 2\sqrt{3} \quad \text{solution}\\\\ x = \dfrac{ \sqrt{ 4y^2+6\sqrt{3}y+9 } } {5} \\\\ x = \dfrac{ \sqrt{ 4 (2\sqrt{3})^2+6\sqrt{3}( 2\sqrt{3})+9 } } {5} \\\\ x = \dfrac{ \sqrt{ 48 +36+9 } } {5} \\\\ x = 1.92873015220\\\\ \mathbf{ \overline{BC} =5x = 9.64365076099 } \end{array} }$$

heureka Jun 24, 2015
#2
+13

@heureka/  Your calculations are so impressive!!!

Jun 24, 2015
#3
+1696
+11

@heureka:/

Jun 24, 2015
#4
+117762
+10

Great work Heureka

Jun 25, 2015
#5
+583
0

Great job,Heureka!It,s correct.

Does this question challenge you?

Jun 25, 2015
#6
+26319
+5

Hallo fiora,

yes, this question challange me.

Thank you for this question!

Jun 25, 2015