In the figure, given below, ABC is an isosceles triangle with BC = 8 cm and AB = AC = 5 cm. Find :

(i) sin B

(ii) tan C.

(iii) sin² B + cos² B

(iv) tan C - cot B

In isosceles Δ ABC, the perpendicular drawn from angle A to the side BC divides BC into 2 equal parts.

BD = DC = $\dfrac{8}{2}$ = 4

In Δ ABD,

⇒ AB² = BD² + AD² (∵ AB is hypotenuse)

⇒ 5² = 4² + AD²

⇒ 25 = 16 + AD²

⇒ AD² = 25 - 16

⇒ AD² = 9

⇒ AD = $\sqrt{9}$

⇒ AD = 3

(i) sin B = $\dfrac{Perpendicular}{Hypotenuse}$

$\dfrac{AD}{AB} = \dfrac{3}{5}$

Hence, sin B = $\dfrac{3}{5}$

(ii) tan C = $\dfrac{Perpendicular}{Base}$

$\dfrac{AD}{DC} = \dfrac{3}{4}$

Hence, tan C = $\dfrac{3}{4}$

(iii) sin² B + cos² B

sin B = $\dfrac{Perpendicular}{Hypotenuse}$

$\dfrac{AD}{AB} = \dfrac{3}{5}$

cos B = $\dfrac{Base}{Hypotenuse}$

$\dfrac{BD}{AB} = \dfrac{4}{5}$

Now, sin² B + cos² B

$= \Big(\dfrac{3}{5}\Big)^2 + \Big(\dfrac{4}{5}\Big)^2\\[1em] = \dfrac{9}{25} + \dfrac{16}{25}\\[1em] = \dfrac{9 + 16}{25}\\[1em] = \dfrac{25}{25}\\[1em] = 1$

Hence, sin² B + cos² B = 1.

(iv) tan C - cot B

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

$= \dfrac{AD}{DC} = \dfrac{3}{4}$

cot B = $\dfrac{Base}{Perpendicular}$

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

Now, tan C - cot B

$= \dfrac{3}{4} - \dfrac{4}{3}\\[1em] = \dfrac{3 \times 3}{4 \times 3}- \dfrac{4 \times 4}{3 \times 4}\\[1em] = \dfrac{9}{12} - \dfrac{16}{12}\\[1em] = \dfrac{9 - 16}{12}\\[1em] = \dfrac{-7}{12}$

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