Question

In a ∆ABC, right-angled at B, it is given that AB = 12 cm and BC = 5 cm.

Find the value of
(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, base = AB and perpendicular = BC
(i) cos A = $\frac{AB}{AC}=\frac{12}{13}$
(ii) cosec A = $\frac{1}{\sin A}=\frac{AC}{BC}=\frac{13}{5}$

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