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Mathematics 7 Online
OpenStudy (anonymous):

In triangle ABC, angle A is a right angle, and angle B = 45 degrees. What is the length of BC? If your answer is not an integer, leave it in simplest form.

OpenStudy (anonymous):

OpenStudy (whpalmer4):

What do you know about this triangle?

OpenStudy (anonymous):

A. 24 ft B. 12 radical 3 ft C. 12 radical 2 ft D. 12 ft

OpenStudy (whpalmer4):

No, don't need the answers, we can find them ourselves. What do you know about angle C in this triangle?

OpenStudy (anonymous):

I don't understand what you are asking I am not good in Geometry I am failing and I need to pass

OpenStudy (anonymous):

is this trigonometry?

OpenStudy (anonymous):

Well I'm in Geometry so I'm guessing it's Geometry

OpenStudy (anonymous):

Have you learnt the Law of sines yet?

OpenStudy (anonymous):

no

OpenStudy (anonymous):

@whpalmer4 How do we teach webbbiee9617 this...uuhh...trig?

OpenStudy (anonymous):

please help me

OpenStudy (anonymous):

first search up "law of sines" and ask me anything you don't understand, ok?

OpenStudy (anonymous):

I searched it ... but how do I set up the problem ?

OpenStudy (whpalmer4):

Okay, no need for law of sines here. The Pythagorean theorem will do the job. Again, do we know anything about angle C? We know angle A and angle B, right? Can we find angle C from just that?

OpenStudy (anonymous):

I did a^2 + b^2 and I got 2025

OpenStudy (anonymous):

how do you find line AB then?

OpenStudy (anonymous):

I'm not sure

OpenStudy (anonymous):

@mathmale how do you teach someone this...uuh...geometry without them knowing law of sines?

OpenStudy (anonymous):

I don't know how to do this ):

OpenStudy (whpalmer4):

What did you use for a and b?

OpenStudy (anonymous):

I used 90 for a and 45 for b

OpenStudy (whpalmer4):

those are angles, not side lengths!

OpenStudy (anonymous):

I only have one side length and that is 12

OpenStudy (anonymous):

now do you agree that we have to teach him the law?

OpenStudy (anonymous):

I am a girl (:

OpenStudy (anonymous):

I'm sorry... I didn't know due to the unknown pic and name

OpenStudy (anonymous):

its fine (: so can somebody please show me how to solve this probelm

OpenStudy (anonymous):

what grade are you in and what formulas have you learnt because I find it odd that they would give you a question that needs something that you haven't learn it

OpenStudy (whpalmer4):

okay, you have a right triangle. You know one leg, and you know two angles. Can you find the other angle? Hint: the angles of a triangle always add up to 180 degrees.

OpenStudy (whpalmer4):

Sorry for the interruption, but someone needed to make a video call on my computer...

OpenStudy (whpalmer4):

Well, you appear to have left. I'll outline what you need to do. We know one leg of the right triangle. We can find the other leg easily, because we know two of the three angles of the triangle are 45 and 90 degrees. That means the third angle must be 180-90-45 = 45 degrees. We have an isosceles triangle. If two angles are equal, the associated sides must also be equal, so the two legs of our triangle are both 12. So, to find the length of the hypotenuse, which is BC, use the Pythagorean theorem, which says that the square of the length of the hypotenuse is the sum of the squares of lengths of the other two legs. If our legs are \(a\) and \(b\) respectively, and the hypotenuse is \(c\), the following equation will be true: \[c^2 = a^2+b^2\] Note that we have \(a = b = 12\). Also, do not make the very frequent mistake of adding \(a^2\) and \(b^2\) and then forgetting to take the square root of the sum to find the value of \(c\).

OpenStudy (anonymous):

im still here had go do dishes

OpenStudy (anonymous):

I got 16.97

OpenStudy (whpalmer4):

That's the correct numeric value, but not the form the answers are looking for. However, I know you had the right number for \(c^2\) :-) \[c^2= 288\] \[c = \sqrt{288}\]To simplify that square root, we can write it as \[\sqrt{288} = \sqrt{2*12^2}\]because we got it by adding \(12^2+12^2\) \[\sqrt{2*12^2} = \sqrt{2}*\sqrt{12^2} = 12*\sqrt{2}\]

OpenStudy (anonymous):

so its 12 radical 2 ?

OpenStudy (whpalmer4):

yes!

OpenStudy (anonymous):

thank you (: I have posted another question

OpenStudy (whpalmer4):

In general, when you have a right isosceles triangle (45/45/90 degrees), the hypotenuse is always in the ratio of \(1:\sqrt{2}\) with the sides. Knowing that, you could simply get the answer for this problem by multiplying \(12*\sqrt{2}\)

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