Asked by sindi
if 13 cos a=12, find the the value of sec a+sina?
Answers
Answered by
Reiny
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
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
Answered by
Hellen mngomezulu
I think that case1 is the one, even though I am not sure math can be sometimes hard
Answered by
Hellen mngomezulu
Case1
Answered by
Kshitija
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
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
Answered by
Kshitija
★ CONTINUE ★
= 5/13
(ii) Sec A = hypo / adj
= AC/ AB
= 13/12
= 5/13
(ii) Sec A = hypo / adj
= AC/ AB
= 13/12
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