Question

In a ∆ABC, ∠B = 90°, AB = 12 cm and BC = 5 cm.

Find:
(i) cos A
(ii) cosec A
(iii) cos C
(iv) coses C.

Answer 1

Using Pythagoras theorem, we get:
AC² = AB² + BC²
⇒ AC² = 12² + 5² = 144 + 25
⇒ AC² = 169
⇒ AC = 13 cm
Now, for T-Ratios of ∠A, AB is base and BC is perpendicular
(i) cosA = $\frac{\mathrm{AB}}{\mathrm{AC}}=\frac{12}{13}$
(ii) cosecA = $\frac{1}{\mathrm{sinA}}=\frac{\mathrm{AC}}{\mathrm{BC}}=\frac{13}{5}$

Similarly, for T-Ratios of ∠C, base = BC and perpendicular = AB
(iii) cosC = $\frac{\mathrm{BC}}{\mathrm{AC}}=\frac{5}{13}$
(iv) cosecC = $\frac{1}{\mathrm{sinC}}=\frac{\mathrm{AC}}{\mathrm{AB}}=\frac{13}{12}$