Find the 10

Find the value of t if 2tx+5y-6=0 and 5x-4y+8=0 are perpendicular, parallel, what angle does each of the lines make with the x-axis, find the angle between the lines?

Correct answer:

t₁ =  2
A₁ =  -38.6598 °
A₂ =  -128.6598 °
t₂ =  -3.125
B₁ =  51.3402 °
B₂ =  -128.6598 °

Step-by-step explanation:

$$\begin{gathered} 2t_x+5y-6=0 \\ 5x-4y+8=0 \\ {n}_1=(2t;5) \\ {n}_2=(5;-4) \\ normal \\ {n}_1\mathrm{.}{n}_2=0 \\ 2\cdot t_1\cdot 5+5\cdot (-4)=0 \\ 10t_1=20 \\ {t}_1=2 \end{gathered}$$

$A_1={\displaystyle \frac{180^{\circ }}{\pi }}\cdot \arctan {\displaystyle \frac{5}{2\cdot t_1}}-90={\displaystyle \frac{180^{\circ }}{\pi }}\cdot \arctan {\displaystyle \frac{5}{2\cdot 2}}-90=-38.6598^{\circ }=-38^{\circ }39^{\mathrm{\prime }}35\mathrm{"}$

$A_2={\displaystyle \frac{180^{\circ }}{\pi }}\cdot \arctan {\displaystyle \frac{-4}{5}}-90=-128.6598^{\circ }=-128^{\circ }39^{\mathrm{\prime }}35\mathrm{"}$

$$\begin{gathered} parallel \\ {n}_1=k\cdot n_2 \\ k=5\mathrm{/}(-4)=-{\displaystyle \frac{5}{4}}=-1\frac{1}{4}=-1.25 \\ 2\cdot t_2=5\mathrm{/}(-4)\cdot 5 \\ 2t_2=-6.25 \\ {t}_2={\displaystyle \frac{-25}{8}}=-3.125=-{\displaystyle \frac{25}{8}} \end{gathered}$$

$B_1={\displaystyle \frac{180^{\circ }}{\pi }}\cdot \arctan {\displaystyle \frac{5}{2\cdot t_2}}-90={\displaystyle \frac{180^{\circ }}{\pi }}\cdot \arctan {\displaystyle \frac{5}{2\cdot (-3.125)}}-90=51.3402^{\circ }=51^{\circ }20^{\mathrm{\prime }}25\mathrm{"}$

$B_2={\displaystyle \frac{180^{\circ }}{\pi }}\cdot \arctan {\displaystyle \frac{-4}{5}}-90=-128.6598^{\circ }=-128^{\circ }39^{\mathrm{\prime }}35\mathrm{"}$

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