# Thread: Trig / right angled triangles / SIN COS TAN

1. ## Trig / right angled triangles / SIN COS TAN

Hi all,
My first post here so my apologies in advance if this is not the correct place to post this.

I have a right angled triangle with one angle of 25˚ (so clearly the last is 65˚). The hypotenuse is 17cm.

My question is how do I work out the other two sides please?

I'm guessing I have to use SIN and have found that:

SIN(A) = opposite(o)/hypotenuse(h)
SIN(65˚) = o/17
SIN(65˚) x 17 = o
0.8268 x 17 = o
o = 14.05

Am I on the right track?

Any help greatly appreciated.

2. Originally Posted by cyberdyne
Hi all,
My first post here so my apologies in advance if this is not the correct place to post this.

I have a right angled triangle with one angle of 25˚ (so clearly the last is 65˚). The hypotenuse is 17cm.

My question is how do I work out the other two sides please?

I'm guessing I have to use SIN and have found that:

SIN(A) = opposite(o)/hypotenuse(h)
SIN(65˚) = o/17
SIN(65˚) x 17 = o
0.8268 x 17 = o
o = 14.05

Am I on the right track?

Any help greatly appreciated.
Looks OK to me, especially since you're not given which angle is 25 degrees.

To find the last side use Pythagoras' Theorem

edit: What masters' said, especially since I didn't check the arithmetic

3. Originally Posted by cyberdyne
Hi all,
My first post here so my apologies in advance if this is not the correct place to post this.

I have a right angled triangle with one angle of 25˚ (so clearly the last is 65˚). The hypotenuse is 17cm.

My question is how do I work out the other two sides please?

I'm guessing I have to use SIN and have found that:

SIN(A) = opposite(o)/hypotenuse(h)
SIN(65˚) = o/17
SIN(65˚) x 17 = o
0.8268 x 17 = o
o = 14.05

Am I on the right track?

Any help greatly appreciated.
Hi cyberdyne,

Yes, you're on the right track except sin (65) = .9063.

4. Originally Posted by cyberdyne
Hi all,
My first post here so my apologies in advance if this is not the correct place to post this.

I have a right angled triangle with one angle of 25˚ (so clearly the last is 65˚). The hypotenuse is 17cm.

My question is how do I work out the other two sides please?

I'm guessing I have to use SIN and have found that:

SIN(A) = opposite(o)/hypotenuse(h)
SIN(65˚) = o/17
SIN(65˚) x 17 = o
0.8268 x 17 = o
o = 14.05

Am I on the right track?

Any help greatly appreciated.
Hi cyberdyne,

welcome to MHF.

$sin55.77^o=0.8267$

$sin65^o=0.9063$

$sin25^o=0.422$

Then you can calculate the lengths of the sides opposite those angles as you were doing

Of course you could also use $cos65^o$ instead of $sin25^o$ for the third side

$sin65^o=\frac{opp}{hyp}=\frac{opp}{17}$

$17sin65^o=opp$

calculates the length of the side opposite the 65 degree angle.

5. Originally Posted by e^(i*pi)
Looks OK to me, especially since you're not given which angle is 25 degrees.

To find the last side use Pythagoras' Theorem
Wow, thanks for the quick reply.

So;
a² = 17² - o²
a² = 289 - 197.4025
a² = 91.5975
a = sq rt of 91.5975
a = 9.57065828
a = 9.57

(it's been a long time since I did these)

6. Originally Posted by Archie Meade
Hi cyberdyne,

welcome to MHF.

$sin65^o=0.9063$
Originally Posted by masters
Hi cyberdyne,

Yes, you're on the right track except sin (65) = .9063.
Ah, OK, thank you. Not sure where I got my calculation from.
And thank you for the welcome.

7. OK so I have:

c=17

SIN(A) = opposite(b)/hypotenuse(h)
SIN(65˚) = b/17
b = SIN(65˚) x 17
b = .9063 x 17
b = 15.4071

c² = a² + b²
17² = a² + 15.4071²
a² = 17² - 15.4071²
a² = 289 - 237.37873
a² = 51.62127
a = sqrt 51.62127
a = 7.184

so a=7.184, b=15.4071, c=17

7.184² + 15.4071² = 288.988586

Presumably you cannot always obtain three exact figures?
Thanks again.

8. Yes,

you have some "round-off", if you work with a number of significant figures.

$17sin65^o=17cos25^o=7.18451044959$

$17sin25=17cos65^o=15.4072323796$

These are the lengths of the perpendicular sides.

$(7.18451044959)^2+(15.4072323796)^2=289=17^2$

9. I can relax now!

10. A couple of extra questions if I may. My daughter had this question in a mock exam paper and I'm trying my best to help her, having not studied Trig myself for approx. 24 yrs!

1. Can anyone please explain how I would find the SIN(E) of 65 degrees on a scientifc calculator. My error above was from entering sin(65) into Google [0.826828679] instead of sin(65 degrees) [0.906307787] but I cannot seem to get the correct value on a calc. It is a relatively basic Texet FX1000
2. Does this question seem a little advanced for a 14 yr old to anyone else? I can't recall using COS, SIN & TAN until my 5th yr of secondary school (todays' yr 11) but my daughter is currently in yr 9.

Many thanks again.

EDIT:
Got it. For some unknown reason I had to hit the RESET button on my calc.
Thanks

11. Originally Posted by cyberdyne
A couple of extra questions if I may. My daughter had this question in a mock exam paper and I'm trying my best to help her, having not studied Trig myself for approx. 24 yrs!

1. Can anyone please explain how I would find the SIN(E) of 65 degrees on a scientifc calculator. My error above was from entering sin(65) into Google [0.826828679] instead of sin(65 degrees) [0.906307787] but I cannot seem to get the correct value on a calc. It is a relatively basic Texet FX1000
2. Does this question seem a little advanced for a 14 yr old to anyone else? I can't recall using COS, SIN & TAN until my 5th yr of secondary school (todays' yr 11) but my daughter is currently in yr 9.

Many thanks again.

EDIT:
Got it. For some unknown reason I had to hit the RESET button on my calc.
Thanks
Hi cyberdyne,

when you calculated $sin65$

the calculator was in radian mode.
The online calculator needed to be in degree mode,
so just set your calculator to degree mode using "deg".

12. Originally Posted by Archie Meade
Hi cyberdyne,

when you calculated $sin65$

the calculator was in radian mode.
The online calculator needed to be in degree mode,
so just set your calculator to degree mode using "deg".
Hi Archie,
This is what I realised after reading the same elsewhere, but no matter what I pressed I could not obtain the correct firgure. I have a feeling the calculator was stuck in radian mode for some reason, hence the RESET. It worked though so all is well.
Many thanks