Question

In triangle ABC, AB = AC = 15 cm and BC = 18 cm. Find:

1. cos B
2. sin C
3. tan² B - sec² B + 2

Answer 1

In the isosceles ΔABC, the perpendicular drawn from angle A to the side BC divides the side BC into two equal parts BD = DC = 9 cm

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

⇒ AD² = 15² – 9²

⇒ AD² = 144 

⇒ AD = `sqrt144`

⇒ AD = 12

(i) cos B = `"base"/"hypotenue" = "BD"/"AB" = (9)/(15) = (3)/(5)`

(ii) sin C = `"perpendicular"/"hypotenuse" = "AD"/"AB" = (12)/(15) = (4)/(5)`

(iii) tan B = `"perpendicular"/"base" = "AD"/"BD" = (12)/(9) = (4)/(3)`

sec B = `"hypotenuse"/"base" = "AB"/"BD" = (15)/(9) = (5)/(3)`

Therefore,

tan² B – sec² B + 2

= `(4/3)^2 –  (5/3)^2+2`

= `16/9 - 25/9 + 2`

= `16/9 - 25/9 + (2 xx 9)/(1 xx 9)`

= `(16  –  25 + 18)/(9)`

= `9/9`

= 1