Questions: Use the figure, which shows two lines whose equations are y1=m1 x+b1 and y2=m2 x+b2. Assume that both lines have positive slopes. Derive a formula for the angle between the two lines. Then use your formula to find the angle between the given pair of lines. y=x and y=(1/sqrt(3)) x θ=

Use the figure, which shows two lines whose equations are y1=m1 x+b1 and y2=m2 x+b2. Assume that both lines have positive slopes. Derive a formula for the angle between the two lines. Then use your formula to find the angle between the given pair of lines.
y=x and y=(1/sqrt(3)) x
θ=

Transcript text: Use the figure, which shows two lines whose equations are $y_{1}=m_{1} x+b_{1}$ and $y_{2}=m_{2} x+b_{2}$. Assume that both lines have positive slopes. Derive a formula for the angle between the two lines. Then use your formula to find the angle between the given pair of lines. \[ y=x \text { and } y=\frac{1}{\sqrt{3}} x \] $\theta=$ $\square$。

Solution

Solution Steps

Step 1: Find the slopes of the lines

The given equations are in slope-intercept form, y = mx + b, where m is the slope. For y = x, the slope m₁ = 1. For y = (1/√3)x, the slope m₂ = 1/√3.

Step 2: Derive the formula for the angle between two lines

Let θ be the angle between the two lines. The tangent of the angle between two lines with slopes m₁ and m₂ is given by the formula: tan(θ) = |(m₂ - m₁) / (1 + m₁m₂)|

Step 3: Plug in the slopes and calculate tan(θ)

Substituting m₁ = 1 and m₂ = 1/√3 into the formula: tan(θ) = |(1/√3 - 1) / (1 + (1)(1/√3))| tan(θ) = |(1 - √3)/√3 / (√3 + 1)/√3| tan(θ) = |(1 - √3) / (1 + √3)| Multiplying the numerator and denominator by (√3 - 1): tan(θ) = |(1 - √3)(√3 - 1) / (1 + √3)(√3 - 1)| tan(θ) = |(2√3 - 4) / (3-1)| tan(θ) = |(2√3 - 4) / 2| tan(θ) = |√3 - 2| Since 2 > √3: tan(θ) = 2 - √3

Step 4: Find the angle θ

θ = arctan(2 - √3) θ = 15°