Question

In \( \triangle ABC \), if \( a = 13 \), \( b = 14 \), and \( c = 15 \), then find the values of:
(i) \( \sec A \)
(ii) \( \csc \frac{A}{2} \)

Hint:

To find the secant and cosecant, use the cosine rule to find angles and apply the appropriate trigonometric identities.

Answer 1

Step 1: Use the cosine rule to find angle \( A \).
By the cosine rule: \[ \cos A = \frac{b^2 + c^2 - a^2}{2bc} \] Substitute the given values: \[ \cos A = \frac{14^2 + 15^2 - 13^2}{2 \times 14 \times 15} = \frac{196 + 225 - 169}{420} = \frac{252}{420} = \frac{3}{5} \]

Step 2: Find \( \sec A \).
Since \( \sec A = \frac{1}{\cos A} \), we have: \[ \sec A = \frac{1}{\frac{3}{5}} = \frac{5}{3} \]

Step 3: Use the sine rule to find angle \( A/2 \).
Using the sine rule for angle \( \frac{A}{2} \), we apply the half angle formula: \[ \sin \frac{A}{2} = \sqrt{\frac{1 - \cos A}{2}} \] Substitute \( \cos A = \frac{3}{5} \): \[ \sin \frac{A}{2} = \sqrt{\frac{1 - \frac{3}{5}}{2}} = \sqrt{\frac{\frac{2}{5}}{2}} = \sqrt{\frac{1}{5}} = \frac{1}{\sqrt{5}} \]

Step 4: Find \( \csc \frac{A}{2} \).
Since \( \csc \frac{A}{2} = \frac{1}{\sin \frac{A}{2}} \), we have: \[ \csc \frac{A}{2} = \frac{1}{\frac{1}{\sqrt{5}}} = \sqrt{5} \]

Final Answer: (i) \( \sec A = \boxed{\frac{5}{3}} \) (ii) \( \csc \frac{A}{2} = \boxed{\sqrt{5}} \)