In the figure (1) given below, ∆ABC is isosceles with AB = AC = 5 cm and BC = 6 cm. Find

(i) sin C

(ii) tan B

(iii) tan C - cot B.

Draw AD perpendicular to BC.

D is the mid point of BC [∵ Perpendicular drawn to the unequal side of an isosceles triangle from the apex vertex bisects the side]

So, BD = CD = 3 cm.

In right-angled ∆ABD,

Using pythagoras theorem we get :

⇒ AB² = AD² + BD²

⇒ AD² = AB² - BD²

⇒ AD² = 5² - 3²

⇒ AD² = 25 - 9

⇒ AD² = 16

⇒ AD = $\sqrt{16}$

⇒ AD = 4 cm.

(i) In right-angled ∆ACD,

sin C = $\dfrac{\text{Perpendicular}}{\text{Hypotenuse}}$

= $\dfrac{AD}{AC} = \dfrac{4}{5}$.

Hence, sin C = $\dfrac{4}{5}$.

(ii) In right-angled ∆ABD,

tan B = $\dfrac{\text{Perpendicular}}{\text{Base}}$

= $\dfrac{AD}{BD} = \dfrac{4}{3}$.

Hence, tan B = $\dfrac{4}{3}$.

(iii) In right-angled ∆ACD,

tan C = $\dfrac{\text{Perpendicular}}{\text{Base}}$

= $\dfrac{AD}{CD} = \dfrac{4}{3}$.

In right-angled ∆ABD,

cot B = $\dfrac{\text{base}}{\text{Perpendicular}}$

= $\dfrac{BD}{AD} = \dfrac{3}{4}$.

Substituting values in tan C - cot B we get :

$\text{tan C} - \text{cot B} = \dfrac{4}{3} - \dfrac{3}{4} \\[1em] = \dfrac{16 - 9}{12} \\[1em] = \dfrac{7}{12}.$

Hence, tan C - cot B = $\dfrac{7}{12}$.