Looks pretty straightforward enough.Can someone please help solve this and show me how come about. This is an assignment but am yet to understand this topic: Thanks
That question is:
find the trigonometric ratios indicated
View attachment 29848
I think you are saying that you have just begun studying trigonometric functions in a right triangle, and don't understand the definitions. In order to help, we need to see an attempt from you, either on part of this, or on a simpler example.Can someone please help solve this and show me how come about. This is an assignment but am yet to understand this topic
Yes am aware of such as sin(Θ)= a/b category and your reference link was actually perfectI think you are saying that you have just begun studying trigonometric functions in a right triangle, and don't understand the definitions. In order to help, we need to see an attempt from you, either on part of this, or on a simpler example.
It may help if you understand that the answers will just be ratios of letters; for example, [imath]\sin(\phi)=\frac{a}{b}[/imath].
For the basics, see
So you have completed this part of the assignment and you do not have any questions regarding this.Yes am aware of
Yes am aware of such as sin(Θ)= a/b category and your reference link was actually perfect
no i don't, this diagram just seem complicated sirSo you have completed this part of the assignment and you do not have any questions regarding this.
Okay so you saw in the reference web-siteno i don't, this diagram just seem complicated sir
Correct.
i got stuck there as there are two switchesOkay so you saw in the reference web-site
sin(Θ) = \(\displaystyle \frac{opposite}{hypotenuse} \)
In your assignment when you are
calculating sin(α)
opposite = ?hypotenuse = ?
Those are not "switches". I interpret those as two boxes. You should type in 107 in the top box and 277 in the bottom boxi got stuck there as there are two switches
You sayView attachment 29858
I do not understand the change here and e.g, sec(0) was suppose to be opposite/Hypotenuse and the (=1/Cos is another change)
You say
................ e.g, sec(Θ) was suppose to be opposite/Hypotenuse
Why do you think so?
You have
cos(Θ) = \(\displaystyle \frac{Adjacent}{Hypotenuse} \)....................... You learnt that from the referenced web-site. Now:
sec(Θ) = \(\displaystyle \frac{1}{cos(\theta)}\) = \(\displaystyle \frac{Hypotenuse}{Adjacent} \)
correct?.................
as we have SOHCAHTOA formula for deriving the sin cos tan.You say
................ e.g, sec(Θ) was suppose to be opposite/Hypotenuse
Why do you think so?
You have
cos(Θ) = \(\displaystyle \frac{Adjacent}{Hypotenuse} \)....................... You learnt that from the referenced web-site. Now:
sec(Θ) = \(\displaystyle \frac{1}{cos(\theta)}\) = \(\displaystyle \frac{Hypotenuse}{Adjacent} \)
correct?.................
How is Sec related to Cos that Sec now have to be the inverse of Cos?
It isn't clear what you are thinking.as we have SOHCAHTOA formula for deriving the sin cos tan.
Is there any formula easy formula to use in deriving the likes of:
View attachment 29861
Tutor have gotten to Trigonometric Ration of special angles. Everyone rushing the cos just to get out and not teaching students. Some move like jet and several Math class are thought at same hour hence the reason i find this here clearer than the tutors
Hi ugudansam. Yes, those are correct.may i confirm if i got the first row of my assignment correctly
oh thank you but why am i not wrong with the sec and cot when i did not put --- Sec0 =z/w (=1/Cos0) ?Hi ugudansam. Yes, those are correct.
However, I think the adjacent side is labeled ω (lower-case Greek letter omega), instead of ε (lower-case Greek letter epsilon).
cos(θ) = ω/Z
sec(θ) = Z/ω
cot(θ) = ω/a
?