if 13 cos a=12, find the the value of sec a+sina?

Answer 1

cosa = 12/13
construct a right-angled triangle and use Pythagoras to find the sides 5, 12, and 13

since the cosine is + in I or IV, we have two cases:

case1: in quad I
sina = 5/13, cosa = 12/13

seca + sina
= 13/12 + 5/13
= 229/156

case2: in quad IV
sina = -5/13, cosa = 12/13
seca + sina
= 13/12 - 5/13
= 109/156

Answer 2

I think that case1 is the one, even though I am not sure math can be sometimes hard

Answer 3

Case1

Answer 4

Let , 13 cos A = 12
Cos A = adjacent/ hypo
= 12/13

AB = 12 , AC = 13

By using Pythagoras theorem
AC² = AB² + BC²
(13)² = (12)² + BC²
169 = 144 + BC²
169 - 144 = BC²
25 = BC²
BC = 5

(i) Sin A = opposite / hypo
= BC/AC

Answer 5

★ CONTINUE ★

= 5/13

(ii) Sec A = hypo / adj
= AC/ AB
= 13/12