In the given figure, ∠B = 90°, AB = 4 units and BC = 3 units. Find:

(i) sin A

(ii) cos A

(iii) cot A

(iv) sin C

(v) sec C

(vi) tan C

In triangle ABC,

By pythagoras theorem,

AC² = AB² + BC²

AC² = 4² + 3²

AC² = 16 + 9

AC² = 25

AC = $\sqrt{25}$

AC = 5 units

(i) sin A = $\dfrac{\text{perpendicular}}{\text{hypotenuse}} = \dfrac{BC}{AC} = \dfrac{3}{5}$

(ii) cos A = $\dfrac{\text{base}}{\text{hypotenuse}} = \dfrac{AB}{AC} = \dfrac{4}{5}$

(iii) cot A = $\dfrac{\text{base}}{\text{perpendicular}} = \dfrac{AB}{BC} = \dfrac{4}{3}$

(iv) sin C = $\dfrac{\text{perpendicular}}{\text{hypotenuse}} = \dfrac{AB}{AC} = \dfrac{4}{5}$

(v) sec C = $\dfrac{\text{hypotenuse}}{\text{base}} = \dfrac{AC}{BC} = \dfrac{5}{3}$

(vi) tan C = $\dfrac{\text{perpendicular}}{\text{base}} = \dfrac{AB}{BC} = \dfrac{4}{3}$