Ask your own question, for FREE!
Mathematics 18 Online
OpenStudy (johnnewman):

What is the value of x?https://static.k12.com/calms_media/media/1422500_1423000/1422767/1/52235abcda5a63ec3d6e60069a91f82606ac9fe4/MS_PA_131002_171216.jpg a. 4 b. 4 of the square root of3 c. 8 d. 8 of the square root of 3

OpenStudy (cderaalien):

Are you studying the Pythagorean theorem ?

OpenStudy (johnnewman):

yes

OpenStudy (cderaalien):

Awesome, okay let's get to solving this.

OpenStudy (johnnewman):

i know a^2+ b^2 =C^2 right

OpenStudy (cderaalien):

Yes, correct.

OpenStudy (cderaalien):

Do you know which part we are trying to solve for?

OpenStudy (cderaalien):

In triangles, there is a leg, leg, and the hypotenuse.

OpenStudy (cderaalien):

In this case, we're solving to find the one of the legs.

OpenStudy (johnnewman):

i think it is a 30, 90 and 60

OpenStudy (cderaalien):

Well, then plug in the values into the theorem

OpenStudy (cderaalien):

a^2 + b^2 = c^2

OpenStudy (cderaalien):

does it equal what you thought?

OpenStudy (cderaalien):

I can tell you how to solve it, or do you think you have it?

OpenStudy (johnnewman):

9000^2

OpenStudy (cderaalien):

Was that your answer?

OpenStudy (cderaalien):

...

OpenStudy (johnnewman):

9000^2

OpenStudy (cderaalien):

9000 x 9000 = 81000000. That is way to large.

OpenStudy (johnnewman):

ok

OpenStudy (phi):

This looks like they want you to use 30-60-90 "special triangle" short-cut the idea is the side opposite the 30 degree angle is 1/2 of the hypotenuse the side opposite the 60 deg angle is the short side times sqr(3)

OpenStudy (johnnewman):

i need to find the square root 60 and 1/2

OpenStudy (cderaalien):

@phi when it is a degree can you still use the Pythagorean theorem and plus in 60^2?

OpenStudy (johnnewman):

3600

OpenStudy (phi):

you can use pythagoras, but when they tell you it's a 30-60-90 triangle, there is a faster way. And here they tell you there is a 60 degree angle and a 90, (and that means the 3rd angle is 30). So they probably expect you to use the short way

OpenStudy (johnnewman):

so i just do 30x30

OpenStudy (phi):

To use pythagoras you need to know 2 of the sides. They only give you the hypotenuse. So you have to use 30-60-90 or trigonometry. the 30-60-90 idea is: label the short leg 1/2 of the hypotenuse label the long leg the len of the short leg times sqrt(3)

OpenStudy (cderaalien):

@JohnNewman do you understand?

OpenStudy (johnnewman):

yes

OpenStudy (johnnewman):

thanks

OpenStudy (phi):

what do you get for the short side?

OpenStudy (johnnewman):

so i do 30 1/2 right

OpenStudy (phi):

what is the length of the hypotenuse ? (the side opposite the 90 deg angle, and the longest side of the triangle). It is labeled in your picture

OpenStudy (johnnewman):

16

OpenStudy (phi):

yes. now what is the "short leg" ? (the rule is 1/2 of the hypotenuse)

OpenStudy (johnnewman):

8

OpenStudy (phi):

and the "long leg" (the side opposite the 60 degree angle) is the short leg times sqr(3) what do you get ?

OpenStudy (johnnewman):

so i need to find the square root of 60 right

OpenStudy (phi):

no. The only way we use the angle is to know that we have a special 30-60-90 triangle, and that the "short leg" is opposite the 30 degree angle and the "long leg" is opposite the 60 deg angle. But to find the lengths of the legs, we use the length of the hypotenuse (not the angles)

OpenStudy (phi):

notice you found the short leg is 8 now find the long leg. It is the length of the short leg times sqr(3)

OpenStudy (johnnewman):

so i times 8 to the square root of 3

OpenStudy (phi):

yes. you can write that like this \[ 8 \sqrt{3}\]

OpenStudy (johnnewman):

ok got it thank phi

Can't find your answer? Make a FREE account and ask your own questions, OR help others and earn volunteer hours!

Join our real-time social learning platform and learn together with your friends!
Can't find your answer? Make a FREE account and ask your own questions, OR help others and earn volunteer hours!

Join our real-time social learning platform and learn together with your friends!