in triangle ABC ,right-angled at C, find cos A,tan A and cosec B if sinA =24/25 - Maths - Introduction to Trigonometry

Question

in triangle ABC ,right-angled at C, find cos A,tan A and cosec B
if sinA =24/25

Kapil · Accepted Answer

ABC is a right angled triangle right angled at C = C = 90
Also for a triangle we know that
A + B + C = 180
= A + B + 90 = 180 = A + B = 180 - 90
= A + B = 90
= A = (90 - B) (1)
Given sin A = 24/25
Now sin² A + cos² A = 1
= cos² A = 1 - sin² A = cos² A = 1 -
(24/25)² (Putting value of sin A)
= cos² A = 1 - 576/625 = cos² A = (625 - 576)/625
= cos² A = 49/625 = cos A = √(49/625)
= cos A = ±7/25
tan A = sin A / cos A = (24/25) / (±7/25)
= tan A = (24/25)(± 25/7)
= tan A = ± 24/7
cosec B = cosec(90 - A) (Putting value of B from equation (1)
)
= sec A = 1/cos A
= 1/(±7/25)
= ±25/7