In a Δ ABC, ∠ B = 90°, AB = 12cm and BC = 5cm. Find
i. cosA
ii. cosecA
iii. cosC
iv. cosecC
Δ ABC is a right angled triangle
By Pythagoras theorem, (hypotenuse)2 = (perpendicular)2 + (base)2
∴AC2 = BC2 + AB2
= AC2 = (5)2 + (12)2
= AC2 = 25 + 144
= AC2 = 169
= (13)2
→ AC = 13
i. cosA = AB/AC = 12/13
ii. cosecA = AC/BC = 13/5
iii. cosC = BC/AC = 5/13
iv. cosecC = AC/AB = 13/12
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