Question

From the following figure, find the values of
(i) sin B 
(ii) tan C
(iii) sec² B - tan²B 
(iv) sin²C + cos²C

Answer 1

Given angle  ADB = 90° and ADC = 90°

⇒ AB² = AD² + BD²            ...( AB is hypotenuse in ΔABD)

⇒  13² = AD² + 5²

∴ AD² = 169 – 25 = 144 and AD = 12

⇒  AC² = AD² + DC²        ...( AC is hypotenuse in ΔADC)

⇒ AC² = 12² + 16²

∴ AC² = 144 + 256 = 400 and AC = 20

(i) sin B = `"perpendicular"/"hypotenuse" = "AD"/"AB" = (12)/(13)`

(ii) tan C = `"perpendicular"/"base" = "AD"/"DC" = (12)/(16) = (3)/(4) `

(iii) sec B = `"hypotenuse"/"base" = "AB"/"BD" = (13)/(5)`

tan B = `"perpendicular"/"base" = "AD"/"BD" = (12)/(5) `

sec² B –  tan² B = `(13/5)^2 – (12/5)^2`

= `(169 – 144)/( 25)`

= `(25)/(25)`

= 1

(iv) sin C = `"perpendicular"/"hypotenuse" = "AD"/"AC" = (12)/(20) = (3)/(5)`

cos C = `"base"/"hypotenuse" = "DC"/"AC" = (16)/(20) = (4)/(5)`

sin² C + cos² C = `(3/5)^2 + (4/5)^2`

= `(9 + 16 )/(25)`

= `(25)/(25)`

= 1