# Trig / right angled triangles / SIN COS TAN

• May 7th 2010, 12:06 PM
cyberdyne
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.
• May 7th 2010, 12:10 PM
e^(i*pi)
Quote:

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
• May 7th 2010, 12:12 PM
masters
Quote:

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.
• May 7th 2010, 12:18 PM
Quote:

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.
• May 7th 2010, 12:19 PM
cyberdyne
Quote:

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) (Headbang)
• May 7th 2010, 12:22 PM
cyberdyne
Quote:

Hi cyberdyne,

welcome to MHF.

$sin65^o=0.9063$

Quote:

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.
• May 7th 2010, 12:37 PM
cyberdyne
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.
• May 7th 2010, 03:53 PM
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.

$(7.18451044959)^2+(15.4072323796)^2=289=17^2$
• May 7th 2010, 11:28 PM
cyberdyne
I can relax now!
• May 8th 2010, 03:12 AM
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
• May 8th 2010, 04:17 AM
Quote:

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".
• May 8th 2010, 04:20 AM
cyberdyne
Quote:

when you calculated $sin65$