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.
Thank you in advance.
OK so I have:
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?
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!
- 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
- 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.
Got it. For some unknown reason I had to hit the RESET button on my calc.
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.