You are using an out of date browser. It may not display this or other websites correctly.

You should upgrade or use an alternative browser.

You should upgrade or use an alternative browser.

- Thread starter ugu
- Start date

- Joined
- Apr 29, 2014

- Messages
- 471

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

Just plug and chug.

- Joined
- Nov 12, 2017

- Messages
- 12,953

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

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

www.mathsisfun.com

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

## Sine, Cosine, Tangent

www.mathsisfun.com

- Joined
- Jun 18, 2007

- Messages
- 25,892

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.

- Joined
- Jun 18, 2007

- Messages
- 25,892

Okay so you saw in the reference web-siteno i don't, this diagram just seem complicated sir

sin(Θ) = \(\displaystyle \frac{opposite}{hypotenuse} \)

In your assignment when you are

calculating sin(α)

opposite = ?

hypotenuse = ?

- Joined
- Jun 18, 2007

- Messages
- 25,892

Correct.

Now calculate the next parameter cos(Θ)

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 = ?

- Joined
- Jun 18, 2007

- Messages
- 25,892

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

- Joined
- Jun 18, 2007

- Messages
- 25,892

You sayView attachment 29858

I do not understand the change here and e.g, sec(0) was suppose to beopposite/Hypotenuseand the (=1/Cosis another change)

................ e.g, sec(Θ) was suppose to be

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?.................

You say

................ e.g, sec(Θ) was suppose to beopposite/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?

as we have SOHCAHTOA formula for deriving the sin cos tan.You say

................ e.g, sec(Θ) was suppose to beopposite/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?.................

Is there any formula easy formula to use in deriving the likes of:

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

- Joined
- Nov 12, 2017

- Messages
- 12,953

How is Sec related to Costhat 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 derivingthe 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

Are you thinking that the

They simply are defined as they are; for some reason, historically, what we call the secant was long ago thought of as more fundamental than what we call the cosine, so it was given a name of its own, while the cosecant was given a name derived from it.

This is a matter of

If that is not what you were thinking, please explain better. Maybe, since you describe "SOHCAHTOA" as "

But in every case, the "co-function" is the function of the complement: coX(theta) = X(90 - theta). That is the part where the names are consistent.

Last edited:

- Joined
- Apr 22, 2015

- Messages
- 3,811

Hi ugudansam. Yes, those are correct.may i confirm if i got the first row of my assignment correctly

However, I think the adjacent side is labeled

cos(θ) = ω/Z

sec(θ) = Z/ω

cot(θ) = ω/a

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 labeledω(lower-case Greek letter omega), instead ofε(lower-case Greek letter epsilon).

cos(θ) = ω/Z

sec(θ) = Z/ω

cot(θ) = ω/a

Last edited: