Calculate, correct to one decimal place, the acute angle between the lines 3x - 4y...

FURTHER MATHEMATICS
WAEC 2010

Calculate, correct to one decimal place, the acute angle between the lines 3x - 4y + 5 = 0 and 2x + 3y - 1 = 0.

  • A. 70.6°
  • B. 50.2°
  • C. 39.8°
  • D. 19.4°

Correct Answer: A. 70.6°

Explanation

\(\tan \theta = \frac{m_{1} - m_{2}}{1 - m_{1}m_{2}}\)

\(m_{1} = \text{slope of 1st line } 4y = 3x + 5 \implies y = \frac{3}{4}x + \frac{5}{4}\)

\(m_{1} = \frac{3}{4}\)

\(m_{2} = \text{slope of 2nd line} 3y = 1 - 2x \implies y = \frac{1}{3} - \frac{2}{3}x\)

\(m_{2} = -\frac{2}{3}\)

\(\tan \theta = \frac{\frac{3}{4} - (-\frac{2}{3})}{1 - ((\frac{3}{4})(-\frac{2}{3}))} = \frac{\frac{17}{12}}{\frac{1}{2}}\)

\(\tan \theta = \frac{17}{6}\)

\(\theta \approxeq 70.6°\)



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