Question

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 + cos²B 
(iv) tan C - cot B

Answer 1

Consider the figure below : 

In the isosceles ΔABC, AB =AC = 5cm and BC = 8cm the perpendicular drawn from angle A to the side BC divides the side BC into two equal parts BD = DC = 4 cm

Since ∠ADB = 90°
⇒ AB² + AD² + BD²   ...( AB is hypotenuse in ΔABD)

⇒ AD² = 5² – 4²

∴ AD² = 9 and AD = 3

(i) sin B = `"AD"/"AB" = (3)/(5)`

(ii) tan C = `"AD"/"DC" = (3)/(4)`

(iii) sin B = `"AD"/"AB" = (3)/(5)`

 cos B = `"BD"/"AB" = (4)/(5)`

Therefore
sin² B + cos² B

= `(3/5)^2 + (4/5)^2`

= `(25)/(25)`

= 1

(iv) tan C = `"AD"/"DC" = (3)/(4)`

cot B = `"BD"/"AD" = (4)/(3)`

Therefore
tan C – cot B

= `(3)/(4) – (4)/(3)`

= `– (7)/(12)`