Solve the right triangle? Round decimals to the nearest tenth?
$begingroup$
Image of the triangle There is a right triangle DEF with the adjacent is 12 and the acute angle of D being 25 degrees and E being the right angle... I have to figure out angle F, DF, and DE... I'm honestly not sure on what to do and i'm stressing out.. I may have worded this wrong or missed something but any help would be appreciated, thank you so much.
triangle
edited Feb 6 '17 at 20:16
K Split X
4,28921232
asked Feb 6 '17 at 19:58
dannehdanneh
13
$endgroup$
add a comment |
$begingroup$
Image of the triangle There is a right triangle DEF with the adjacent is 12 and the acute angle of D being 25 degrees and E being the right angle... I have to figure out angle F, DF, and DE... I'm honestly not sure on what to do and i'm stressing out.. I may have worded this wrong or missed something but any help would be appreciated, thank you so much.
triangle
edited Feb 6 '17 at 20:16
K Split X
4,28921232
asked Feb 6 '17 at 19:58
dannehdanneh
13
$endgroup$
$begingroup$
This is basic trig. Draw a picture
$endgroup$
– K Split X
Feb 6 '17 at 20:00
$begingroup$
Check out my answer, if you dont understand something tell me
$endgroup$
– K Split X
Feb 6 '17 at 20:13
add a comment |
$begingroup$
Image of the triangle There is a right triangle DEF with the adjacent is 12 and the acute angle of D being 25 degrees and E being the right angle... I have to figure out angle F, DF, and DE... I'm honestly not sure on what to do and i'm stressing out.. I may have worded this wrong or missed something but any help would be appreciated, thank you so much.
triangle
edited Feb 6 '17 at 20:16
K Split X
4,28921232
asked Feb 6 '17 at 19:58
dannehdanneh
13
$endgroup$
Image of the triangle There is a right triangle DEF with the adjacent is 12 and the acute angle of D being 25 degrees and E being the right angle... I have to figure out angle F, DF, and DE... I'm honestly not sure on what to do and i'm stressing out.. I may have worded this wrong or missed something but any help would be appreciated, thank you so much.
triangle
triangle
edited Feb 6 '17 at 20:16
K Split X
4,28921232
asked Feb 6 '17 at 19:58
dannehdanneh
13
edited Feb 6 '17 at 20:16
K Split X
4,28921232
asked Feb 6 '17 at 19:58
dannehdanneh
13
edited Feb 6 '17 at 20:16
K Split X
4,28921232
edited Feb 6 '17 at 20:16
K Split X
4,28921232
edited Feb 6 '17 at 20:16
K Split X
4,28921232
4,28921232
asked Feb 6 '17 at 19:58
dannehdanneh
13
asked Feb 6 '17 at 19:58
dannehdanneh
13
asked Feb 6 '17 at 19:58
dannehdanneh
13
13
$begingroup$
This is basic trig. Draw a picture
$endgroup$
– K Split X
Feb 6 '17 at 20:00
$begingroup$
Check out my answer, if you dont understand something tell me
$endgroup$
– K Split X
Feb 6 '17 at 20:13
add a comment |
$begingroup$
This is basic trig. Draw a picture
$endgroup$
– K Split X
Feb 6 '17 at 20:00
$begingroup$
Check out my answer, if you dont understand something tell me
$endgroup$
– K Split X
Feb 6 '17 at 20:13
$begingroup$
This is basic trig. Draw a picture
$endgroup$
– K Split X
Feb 6 '17 at 20:00
$begingroup$
This is basic trig. Draw a picture
$endgroup$
– K Split X
Feb 6 '17 at 20:00
$begingroup$
Check out my answer, if you dont understand something tell me
$endgroup$
– K Split X
Feb 6 '17 at 20:13
$begingroup$
Check out my answer, if you dont understand something tell me
$endgroup$
– K Split X
Feb 6 '17 at 20:13
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Angle $F$. Remember, the insides of a triangle add up to $180$ degrees.
Length of $DF$. Remember, $sin theta=frac{opp}{hyp}=frac{12}{DF}$
So we have $sin 25=frac{12}{DF}$. You can solve from here.
Length of $DE$. After you calculate angle $F$, then we know $tantheta=frac{opp}{adj}=frac{DE}{9}$
So we have $tan text{ (angle F)}=frac{DE}{9}$. You can solve from here.
answered Feb 6 '17 at 20:07
K Split XK Split X
4,28921232
$endgroup$
$begingroup$
I'm not being lazy, but i'm honestly having a hard time understanding this whole section. How do i figure out the sin 25= 12/DF? It makes no sense to me and i'm feeling stupid here. Also, the same with the length of DE. I just don't get any of this and its frustrating
$endgroup$
– danneh
Feb 6 '17 at 20:19
$begingroup$
sin $25$ you just evaluate in your calculator. We basically have $sin(25) = frac{12}{DF}$, and so solving for $DF$, we see that $DF=frac{12}{sin(25)} = 28.4$
$endgroup$
– K Split X
Feb 6 '17 at 20:35
$begingroup$
If you don't understand your textbook, I highly recommend khanacademy.
$endgroup$
– K Split X
Feb 6 '17 at 20:36
$begingroup$
youtube.com/watch?v=Jsiy4TxgIME
$endgroup$
– K Split X
Feb 6 '17 at 20:36
$begingroup$
is solving for DE the same as we solved for DF?
$endgroup$
– danneh
Feb 6 '17 at 20:38
|
show 3 more comments
Your Answer
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Angle $F$. Remember, the insides of a triangle add up to $180$ degrees.
Length of $DF$. Remember, $sin theta=frac{opp}{hyp}=frac{12}{DF}$
So we have $sin 25=frac{12}{DF}$. You can solve from here.
Length of $DE$. After you calculate angle $F$, then we know $tantheta=frac{opp}{adj}=frac{DE}{9}$
So we have $tan text{ (angle F)}=frac{DE}{9}$. You can solve from here.
answered Feb 6 '17 at 20:07
K Split XK Split X
4,28921232
$endgroup$
$begingroup$
I'm not being lazy, but i'm honestly having a hard time understanding this whole section. How do i figure out the sin 25= 12/DF? It makes no sense to me and i'm feeling stupid here. Also, the same with the length of DE. I just don't get any of this and its frustrating
$endgroup$
– danneh
Feb 6 '17 at 20:19
$begingroup$
sin $25$ you just evaluate in your calculator. We basically have $sin(25) = frac{12}{DF}$, and so solving for $DF$, we see that $DF=frac{12}{sin(25)} = 28.4$
$endgroup$
– K Split X
Feb 6 '17 at 20:35
$begingroup$
If you don't understand your textbook, I highly recommend khanacademy.
$endgroup$
– K Split X
Feb 6 '17 at 20:36
$begingroup$
youtube.com/watch?v=Jsiy4TxgIME
$endgroup$
– K Split X
Feb 6 '17 at 20:36
$begingroup$
is solving for DE the same as we solved for DF?
$endgroup$
– danneh
Feb 6 '17 at 20:38
|
show 3 more comments
$begingroup$
Angle $F$. Remember, the insides of a triangle add up to $180$ degrees.
Length of $DF$. Remember, $sin theta=frac{opp}{hyp}=frac{12}{DF}$
So we have $sin 25=frac{12}{DF}$. You can solve from here.
Length of $DE$. After you calculate angle $F$, then we know $tantheta=frac{opp}{adj}=frac{DE}{9}$
So we have $tan text{ (angle F)}=frac{DE}{9}$. You can solve from here.
answered Feb 6 '17 at 20:07
K Split XK Split X
4,28921232
$endgroup$
$begingroup$
I'm not being lazy, but i'm honestly having a hard time understanding this whole section. How do i figure out the sin 25= 12/DF? It makes no sense to me and i'm feeling stupid here. Also, the same with the length of DE. I just don't get any of this and its frustrating
$endgroup$
– danneh
Feb 6 '17 at 20:19
$begingroup$
sin $25$ you just evaluate in your calculator. We basically have $sin(25) = frac{12}{DF}$, and so solving for $DF$, we see that $DF=frac{12}{sin(25)} = 28.4$
$endgroup$
– K Split X
Feb 6 '17 at 20:35
$begingroup$
If you don't understand your textbook, I highly recommend khanacademy.
$endgroup$
– K Split X
Feb 6 '17 at 20:36
$begingroup$
youtube.com/watch?v=Jsiy4TxgIME
$endgroup$
– K Split X
Feb 6 '17 at 20:36
$begingroup$
is solving for DE the same as we solved for DF?
$endgroup$
– danneh
Feb 6 '17 at 20:38
|
show 3 more comments
$begingroup$
Angle $F$. Remember, the insides of a triangle add up to $180$ degrees.
Length of $DF$. Remember, $sin theta=frac{opp}{hyp}=frac{12}{DF}$
So we have $sin 25=frac{12}{DF}$. You can solve from here.
Length of $DE$. After you calculate angle $F$, then we know $tantheta=frac{opp}{adj}=frac{DE}{9}$
So we have $tan text{ (angle F)}=frac{DE}{9}$. You can solve from here.
answered Feb 6 '17 at 20:07
K Split XK Split X
4,28921232
$endgroup$
Angle $F$. Remember, the insides of a triangle add up to $180$ degrees.
Length of $DF$. Remember, $sin theta=frac{opp}{hyp}=frac{12}{DF}$
So we have $sin 25=frac{12}{DF}$. You can solve from here.
Length of $DE$. After you calculate angle $F$, then we know $tantheta=frac{opp}{adj}=frac{DE}{9}$
So we have $tan text{ (angle F)}=frac{DE}{9}$. You can solve from here.
answered Feb 6 '17 at 20:07
K Split XK Split X
4,28921232
answered Feb 6 '17 at 20:07
K Split XK Split X
4,28921232
answered Feb 6 '17 at 20:07
K Split XK Split X
4,28921232
answered Feb 6 '17 at 20:07
K Split XK Split X
4,28921232
4,28921232
$begingroup$
I'm not being lazy, but i'm honestly having a hard time understanding this whole section. How do i figure out the sin 25= 12/DF? It makes no sense to me and i'm feeling stupid here. Also, the same with the length of DE. I just don't get any of this and its frustrating
$endgroup$
– danneh
Feb 6 '17 at 20:19
$begingroup$
sin $25$ you just evaluate in your calculator. We basically have $sin(25) = frac{12}{DF}$, and so solving for $DF$, we see that $DF=frac{12}{sin(25)} = 28.4$
$endgroup$
– K Split X
Feb 6 '17 at 20:35
$begingroup$
If you don't understand your textbook, I highly recommend khanacademy.
$endgroup$
– K Split X
Feb 6 '17 at 20:36
$begingroup$
youtube.com/watch?v=Jsiy4TxgIME
$endgroup$
– K Split X
Feb 6 '17 at 20:36
$begingroup$
is solving for DE the same as we solved for DF?
$endgroup$
– danneh
Feb 6 '17 at 20:38
|
show 3 more comments
$begingroup$
I'm not being lazy, but i'm honestly having a hard time understanding this whole section. How do i figure out the sin 25= 12/DF? It makes no sense to me and i'm feeling stupid here. Also, the same with the length of DE. I just don't get any of this and its frustrating
$endgroup$
– danneh
Feb 6 '17 at 20:19
$begingroup$
sin $25$ you just evaluate in your calculator. We basically have $sin(25) = frac{12}{DF}$, and so solving for $DF$, we see that $DF=frac{12}{sin(25)} = 28.4$
$endgroup$
– K Split X
Feb 6 '17 at 20:35
$begingroup$
If you don't understand your textbook, I highly recommend khanacademy.
$endgroup$
– K Split X
Feb 6 '17 at 20:36
$begingroup$
youtube.com/watch?v=Jsiy4TxgIME
$endgroup$
– K Split X
Feb 6 '17 at 20:36
$begingroup$
is solving for DE the same as we solved for DF?
$endgroup$
– danneh
Feb 6 '17 at 20:38
$begingroup$
I'm not being lazy, but i'm honestly having a hard time understanding this whole section. How do i figure out the sin 25= 12/DF? It makes no sense to me and i'm feeling stupid here. Also, the same with the length of DE. I just don't get any of this and its frustrating
$endgroup$
– danneh
Feb 6 '17 at 20:19
$begingroup$
I'm not being lazy, but i'm honestly having a hard time understanding this whole section. How do i figure out the sin 25= 12/DF? It makes no sense to me and i'm feeling stupid here. Also, the same with the length of DE. I just don't get any of this and its frustrating
$endgroup$
– danneh
Feb 6 '17 at 20:19
$begingroup$
sin $25$ you just evaluate in your calculator. We basically have $sin(25) = frac{12}{DF}$, and so solving for $DF$, we see that $DF=frac{12}{sin(25)} = 28.4$
$endgroup$
– K Split X
Feb 6 '17 at 20:35
$begingroup$
sin $25$ you just evaluate in your calculator. We basically have $sin(25) = frac{12}{DF}$, and so solving for $DF$, we see that $DF=frac{12}{sin(25)} = 28.4$
$endgroup$
– K Split X
Feb 6 '17 at 20:35
$begingroup$
If you don't understand your textbook, I highly recommend khanacademy.
$endgroup$
– K Split X
Feb 6 '17 at 20:36
$begingroup$
If you don't understand your textbook, I highly recommend khanacademy.
$endgroup$
– K Split X
Feb 6 '17 at 20:36
$begingroup$
youtube.com/watch?v=Jsiy4TxgIME
$endgroup$
– K Split X
Feb 6 '17 at 20:36
$begingroup$
youtube.com/watch?v=Jsiy4TxgIME
$endgroup$
– K Split X
Feb 6 '17 at 20:36
$begingroup$
is solving for DE the same as we solved for DF?
$endgroup$
– danneh
Feb 6 '17 at 20:38
$begingroup$
is solving for DE the same as we solved for DF?
$endgroup$
– danneh
Feb 6 '17 at 20:38
|
show 3 more comments
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$begingroup$
This is basic trig. Draw a picture
$endgroup$
– K Split X
Feb 6 '17 at 20:00
$begingroup$
Check out my answer, if you dont understand something tell me
$endgroup$
– K Split X
Feb 6 '17 at 20:13