Applying Pythagoras theorem for \(ΔABC,\)we obtain
\(\text{AC}^ 2 =\text{ AB}^ 2 + \text{BC}^ 2\)
\(= (24\text{ cm})^2 + (7 \text{cm})^ 2\)
\(= (576 + 49)\)cm2
\(= 625\) cm2
\(∴\) AC = \(\sqrt{625}\) cm = 25 cm
(i)

\(\text{ sin A} = \frac{\text{Side}\ \text{ Opposite}\ \text{ to}\ ∠A }{\text{Hypotenuse}}\) \(= \frac{\text{BC}}{\text{AC}} = \frac{7}{25}\)
\(\text{ Cos A} = \frac{\text{Side}\ \text{ Adjacent}\ \text{ to}\ ∠A }{\text{Hypotenuse}}\)\(= \frac{\text{AB}}{\text{AC}} = \frac{24}{25}\)
(ii) 
\(\text{ sin C} = \frac{\text{Side}\ \text{ Opposite}\ \text{ to}\ ∠C }{\text{Hypotenuse}}\)e \(= \frac{\text{AB}}{\text{AC}} = \frac{24}{25}\)
\(\text{ Cos C} = \frac{\text{Side}\ \text{ Adjacent}\ \text{ to}\ ∠C }{\text{Hypotenuse}}\)\(= \frac{\text{BC}}{\text{AC}} = \frac{7}{25}\)

If sin A = \(\frac{3}{4}\), calculate cos A and tan A.
| Case No. | Lens | Focal Length | Object Distance |
|---|---|---|---|
| 1 | \(A\) | 50 cm | 25 cm |
| 2 | B | 20 cm | 60 cm |
| 3 | C | 15 cm | 30 cm |