In the given figure, AD ⊥ BC.

If AB = 13 cm, BD = 5 cm and DC = 16 cm, find the values of :

(i) sin B

(ii) sec B

(iii) cot B

(iv) cos C

(v) cosec C

(vi) tan C

Using pythagoras theorem in right angled triangle ADB

AB² = BD² + AD²

AD² = AB² - BD²

AD² = 13² - 5²

AD² = 169 - 25

AD² = 144

AD = $\sqrt{144}$

AD = 12 cm

Now we will find out AC using pythagoras theorem in right angled triangle ADC,

AC² = DC² + AD²

AC² = 16² + 12²

AC² = 256 + 144

AC² = 400

AC = $\sqrt{400}$

AC = 20 cm

Now,

(i) sin B = $\dfrac{\text{perpendicular}}{\text{hypotenuse}} = \dfrac{AD}{AB} = \dfrac{12}{13}$

(ii) sec B = $\dfrac{\text{hypotenuse}}{\text{base}} = \dfrac{AB}{BD} = \dfrac{13}{5}$

(iii) cot B = $\dfrac{\text{base}}{\text{perpendicular}} = \dfrac{BD}{AD} = \dfrac{5}{12}$

(iv) cos C = $\dfrac{\text{base}}{\text{hypotenuse}} = \dfrac{DC}{AC} = \dfrac{16}{20} = \dfrac{4}{5}$

(v) cosec C = $\dfrac{\text{hypotenuse}}{\text{perpendicular}} = \dfrac{AC}{AD} = \dfrac{20}{12} = \dfrac{5}{3}$

(vi) tan C = $\dfrac{\text{perpendicular}}{\text{base}} = \dfrac{AD}{DC} = \dfrac{12}{16} = \dfrac{3}{4}$