Question

14 mark for review triangles abc and def are similar triangles. angles a and d are right angles, and angle b corresponds to angle e. if \\(\cos(c) = \frac{23}{265}\\), what is the value of \\(\cos(f)\\)?
a \\(\frac{23}{265}\\)
b \\(\frac{264}{265}\\)
c \\(\frac{265}{264}\\)
d \\(\frac{265}{23}\\)
15 mark for review what is the slope of line \\(a\\) if line \\(a\\) is perpendicular to line \\(b\\), defined as \\(9x = 3y - 6\\)?

Explanation:

Question 14

Step1: Analyze similar triangles' angles

Since \(\triangle ABC \sim \triangle DEF\), corresponding angles are equal. \(\angle A = \angle D = 90^\circ\), \(\angle B = \angle E\), so \(\angle C = \angle F\) (because triangle angles sum to \(180^\circ\), so the third angles must be equal).

Step2: Use cosine of equal angles

If \(\angle C = \angle F\), then \(\cos(C)=\cos(F)\). Given \(\cos(C)=\frac{23}{265}\), so \(\cos(F)=\frac{23}{265}\).

Step1: Find slope of line \(b\)

Rewrite line \(b\)'s equation \(9x = 3y - 6\) in slope - intercept form \(y = mx + c\) (where \(m\) is slope).
Add 6 to both sides: \(3y=9x + 6\).
Divide by 3: \(y = 3x+2\). So slope of line \(b\) (\(m_b\)) is 3.

Step2: Find slope of perpendicular line \(a\)

If two lines are perpendicular, the product of their slopes is \(- 1\). Let slope of line \(a\) be \(m_a\). Then \(m_a\times m_b=-1\).
Substitute \(m_b = 3\): \(m_a\times3=-1\).
Solve for \(m_a\): \(m_a=-\frac{1}{3}\).

Answer:

A. \(\frac{23}{265}\)

Question 15