A(0, 0), B(6,5, 0) and C(10, 7) are three points on a circle - Leaving Cert Mathematics - Question 4 - 2017

To find the angle |∠BCA|, we will determine the slopes of lines AC and BC.

Finding the slopes:

• Slope of AC: $ext{slope of AC} = \frac{y_C - y_A}{x_C - x_A} = \frac{7 - 0}{10 - 0} = \frac{7}{10}$

• Slope of BC: $ext{slope of BC} = \frac{y_C - y_B}{x_C - x_B} = \frac{7 - 5}{10 - 6} = \frac{2}{4} = \frac{1}{2}$

Finding the angle:

Use the formula for the angle between two lines:

$\tan(\theta) = \left| \frac{m_1 - m_2}{1 + m_1m_2} \right|$ where $m_1 = \frac{7}{10}$ and $m_2 = \frac{1}{2}$. Thus,

$\tan(\theta) = \left| \frac{\frac{7}{10} - \frac{1}{2}}{1 + \left(\frac{7}{10}\right)\left(\frac{1}{2}\right)} \right|$

Calculating:

\theta = \tan^{-1}(\frac{2}{27}) \approx 4.24^{\circ}$$ Thus, the angle |∠BCA| is approximately **4.24°**.