Help please?

[toxicmindjunk]

I have this question on my math packet for school and I don't know where to start.

What is the length of the longer leg of a 30-60-90 triangle with a hypotenuse length of 24 ?(3)?

Can someone help me out?
It'd be much appreciated.

 

[galactus]

Use cosine. Let the longer side have length x.

Then, \(\displaystyle cos(\frac{\pi}{6})=\frac{x}{24\sqrt{3}m}\)

Solve for x. Your solution will be in terms of m.

Or you could use secant.

\(\displaystyle sec(\frac{\pi}{6})=\frac{24\sqrt{3}m}{x}\)

Solve for x.

Many ways to go about it.

See why these are. They are the trig laws.

Say we have a right triangle. Let the smaller angle be x (in your case, the 30 degrees or Pi/6 radians. I prefer radians) and let the adjacent side from that angle be a, the opposite side be b and the hypoteneuse be c.

Then, sin(x)=b/c, cos(x)=a/c, tan(x)=b/a, sec(x)=c/a, csc(x)=c/b, cot(x)=a/b.

 

[toxicmindjunk]

Use cosine. Let the longer side have length x.

Then, \(\displaystyle cos(\frac{\pi}{6})=\frac{x}{24\sqrt{3}m}\)

Solve for x. Your solution will be in terms of m.

Or you could use secant.

\(\displaystyle sec(\frac{\pi}{6})=\frac{24\sqrt{3}m}{x}\)

Solve for x.

Many ways to go about it.

See why these are. They are the trig laws.

Say we have a right triangle. Let the smaller angle be x (in your case, the 30 degrees or Pi/6 radians. I prefer radians) and let the adjacent side from that angle be a, the opposite side be b and the hypoteneuse be c.

Then, sin(x)=b/c, cos(x)=a/c, tan(x)=b/a, sec(x)=c/a, csc(x)=c/b, cot(x)=a/b.

Oops. I think you interpreted "m" as a variable when I'm almost certain it's meters. Does it make a difference?

 

[galactus]

Yes, it does make a difference. The way it was written suggested a variable. Anyway, do the same as I posted before, only leave out the m. Then, the results will be in meters.

Like so,

\(\displaystyle cos(\frac{\pi}{6})=\frac{x}{24\sqrt{3}}\)

Solve for x. The solution will be a nice integer.

 

[toxicmindjunk]

Ok, thanks

 

[galactus]

Mind, you can also use degrees if that's what you're working in. If you use your calculator, make sure it is the correct mode. If it is in radian mode and you use degrees, you'll get a wacky answer.

\(\displaystyle cos(30)=\frac{x}{24\sqrt{3}}\).

It is a common mistake to have a calculator in the wrong mode when performing trig operations. Just a heads up. Good luck.

 

[Denis]

Or let hypotenuse = h = 24sqrt(3)
Longer leg = sqrt(h^2 - (h/2)^2) = hsqrt(3)/2

 

[Mrspi]

You should not need to use either trig OR a calculator to answer this question.

I'm assuming you may be studying the 30-60-90 theorem in geometry....you can find a complete explanation here:

http://regentsprep.org/Regents/mathb/5B1/Tri30.htm

You'll see there that the longer leg (LL) in the explanation, is hypotenuse/sqrt(3), or H/sqrt(3) in the explanation.

Your hypotenuse is GIVEN to be 24 sqrt(3)

Substitute 24 sqrt(3) for "H" in " H / sqrt(3)"

[24 sqrt(3)] / sqrt(3)

Simplify the fraction. You should NOT need a calculator for that!

I think that knowing the relationships between the sides and angles of the special right triangles (30-60-90 and 45-45-90) are EXTREMELY important, and save you a ton of grief in calculus.

Yep, yep...I KNOW...you can use a calculator. BUT WHY??????