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Math  /  Trigonometry

QuestionThe line kk has a slope of -2 . The line jj makes an angle of 30∘30^{\circ} with kk. Find one possible value of the slope of the line jj. Give your answer in the form d+efd+e \sqrt{f}, where d,e,f∈Zd, e, f \in \mathbb{Z}.

Studdy Solution

STEP 1

1. The slope of line k k is −2 -2 .
2. Line j j makes an angle of 30∘ 30^\circ with line k k .
3. We are looking for one possible slope of line j j .

STEP 2

1. Recall the formula for the angle between two lines in terms of their slopes.
2. Set up the equation using the given angle and solve for the slope of line j j .
3. Simplify the expression to the desired form.

STEP 3

Recall the formula for the angle θ\theta between two lines with slopes m1m_1 and m2m_2:
tan⁡(θ)=∣m2−m11+m1m2∣\tan(\theta) = \left| \frac{m_2 - m_1}{1 + m_1 m_2} \right|

STEP 4

Set up the equation using the given angle 30∘30^\circ and the slope of line k k , which is −2-2. Let mj m_j be the slope of line j j :
tan⁡(30∘)=∣mj+21−2mj∣\tan(30^\circ) = \left| \frac{m_j + 2}{1 - 2m_j} \right|
Since tan⁡(30∘)=13\tan(30^\circ) = \frac{1}{\sqrt{3}}, we have:
13=∣mj+21−2mj∣\frac{1}{\sqrt{3}} = \left| \frac{m_j + 2}{1 - 2m_j} \right|

STEP 5

Solve the equation:
Case 1:
mj+21−2mj=13\frac{m_j + 2}{1 - 2m_j} = \frac{1}{\sqrt{3}}
Cross-multiply:
3(mj+2)=1−2mj\sqrt{3}(m_j + 2) = 1 - 2m_j
Simplify:
3mj+23=1−2mj\sqrt{3}m_j + 2\sqrt{3} = 1 - 2m_j
Combine like terms:
3mj+2mj=1−23\sqrt{3}m_j + 2m_j = 1 - 2\sqrt{3}
mj(3+2)=1−23m_j(\sqrt{3} + 2) = 1 - 2\sqrt{3}
Solve for mj m_j :
mj=1−233+2m_j = \frac{1 - 2\sqrt{3}}{\sqrt{3} + 2}
Rationalize the denominator:
Multiply numerator and denominator by the conjugate of the denominator:
mj=(1−23)(3−2)(3+2)(3−2)m_j = \frac{(1 - 2\sqrt{3})(\sqrt{3} - 2)}{(\sqrt{3} + 2)(\sqrt{3} - 2)}
Calculate the denominator:
(3+2)(3−2)=3−4=−1(\sqrt{3} + 2)(\sqrt{3} - 2) = 3 - 4 = -1
Calculate the numerator:
(1−23)(3−2)=3−2−2(3)+43=53−8(1 - 2\sqrt{3})(\sqrt{3} - 2) = \sqrt{3} - 2 - 2(3) + 4\sqrt{3} = 5\sqrt{3} - 8
Thus:
mj=53−8−1=8−53m_j = \frac{5\sqrt{3} - 8}{-1} = 8 - 5\sqrt{3}
The slope of line j j is:
8−53 \boxed{8 - 5\sqrt{3}}

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