+15 hey guys and gals I’m Kayla and I’m kinda new to this site but I need a LOT of help on my trig lol, so I decided to ask some questions if that’s okay! Answers are greatly appreciated thanks :)
1. Convert 136(degrees) 14’ 18” to decimal form
2. Convert -22.853(degreees) to DMS form
3. If cos()=1/3 use the trigonometric identities to find tan()
if you you can answer any of these THANK YOU, also don’t be afraid to DM me if you want to chat or anything lol!
+9481 Welcome to the forum!! 
1. 1 ' = 1/60 ° → 14 ' = 14/60 °
1 " = 1/3600 ° → 18 " = 18/3600 °
136 ° + 14 ' + 18 " =
136 ° + 14/60 ° + 18/3600 ° =
81743/600 ° ≈ 136.238°
2. 1 ° = 60 ' → 0.853 ° = 0.853 * 60 ' = 51.18 '
1 ' = 60 " → 0.18 ' = 0.18 * 60 " = 10.8 "
-( 22.853 ° ) =
-( 22 ° + 0.853 ° ) =
-( 22 ° + 51.18 ' ) =
-( 22 ° + 51 ' + 0.18 ' ) =
-( 22 ° + 51 ' + 10.8 " ) = - 22 ° 51 ' 10.8 "
3. cos θ = 1/3
Use the Pythagorean identity to find sin θ .
sin2 θ + cos2 θ = 1
sin2 θ + (1/3)2 = 1
sin2 θ + 1/9 = 1
sin2 θ = 1 - 1/9
sin2 θ = 8/9
sin θ = ±√( 8/9 ) = ± √8 / 3 Use these values for sin and cos in the definition of tan .
tan θ = sin θ / cos θ
tan θ = [ ± √8 / 3 ] / [ 1/3 ] Invert the second fraction and multiply.
tan θ = [ ± √8 / 3 ] * [ 3/1 ]
tan θ = ± √8
tan θ = ± 2√2
If you have a question about one of these, please don't hesitate to ask!
+9481 Welcome to the forum!!
1. 1 ' = 1/60 ° → 14 ' = 14/60 °
1 " = 1/3600 ° → 18 " = 18/3600 °
136 ° + 14 ' + 18 " =
136 ° + 14/60 ° + 18/3600 ° =
81743/600 ° ≈ 136.238°
2. 1 ° = 60 ' → 0.853 ° = 0.853 * 60 ' = 51.18 '
1 ' = 60 " → 0.18 ' = 0.18 * 60 " = 10.8 "
-( 22.853 ° ) =
-( 22 ° + 0.853 ° ) =
-( 22 ° + 51.18 ' ) =
-( 22 ° + 51 ' + 0.18 ' ) =
-( 22 ° + 51 ' + 10.8 " ) = - 22 ° 51 ' 10.8 "
3. cos θ = 1/3
Use the Pythagorean identity to find sin θ .
sin2 θ + cos2 θ = 1
sin2 θ + (1/3)2 = 1
sin2 θ + 1/9 = 1
sin2 θ = 1 - 1/9
sin2 θ = 8/9
sin θ = ±√( 8/9 ) = ± √8 / 3 Use these values for sin and cos in the definition of tan .
tan θ = sin θ / cos θ
tan θ = [ ± √8 / 3 ] / [ 1/3 ] Invert the second fraction and multiply.
tan θ = [ ± √8 / 3 ] * [ 3/1 ]
tan θ = ± √8
tan θ = ± 2√2
If you have a question about one of these, please don't hesitate to ask!
