How ratios are written here? (Circle center O, r=7, m(BOC)=120; |BC|=?)

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Ganesh Ujwal

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In the below figure, O is the center of the circle. Radius is 7 cm. [ tex]\angle BOC[ /tex] = 120 Find the length of BC?

1.jpg

Solution: Join AO to meet BC in M

\(\displaystyle \Delta ABC\) \(\displaystyle \cong\) \(\displaystyle \Delta AOC\) and

\(\displaystyle \Delta BMO\) \(\displaystyle \cong\) \(\displaystyle \Delta CMO\)

\(\displaystyle \angle BMO\) = 90

Triangle BMO has three angles are 90, 60 and 30.

The sides are in the ratio of 2: \(\displaystyle \sqrt 3\) : 1

2X = 7 \(\displaystyle \Rightarrow\) X = 7/2

BM = \(\displaystyle \Rightarrow\) \(\displaystyle \sqrt 3\) x 7 / 2 = 7\(\displaystyle \sqrt 3 / 2\)

BC = 2 x 7\(\displaystyle \sqrt 3 / 2\) = 7 \(\displaystyle \sqrt 3\)
 
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Ganesh Ujwal said:
In the below figure, O is the center of the circle. Radius is 7 cm. $\angle$ BOC = 120 Find the length of BC?

View attachment 9364

Solution: Join AO to meet BC in M

\triangle ABC$ $\cong$ $\triangle AOC$ and

$\triangle$ BMO $\cong$ $\triangle CMO$
$\angle BMO$ = 90

Triangle BMO has three angles are 90, 60 and 30.

The sides are in the ratio of 2: $\sqrt 3$ : 1

2X = 7 $\Rightarrow$ X = 7/2

BM = $\Rightarrow$ $\sqrt 3$ x 7 / 2 = 7$\sqrt 3 / 2$

BC = 2 x 7$\sqrt 3 / 2$ = 7 $\sqrt 3$
Click to expand...
First, do not use dollar signs as delimiters for LaTeX here. Instead use [ tex] and [ /tex] but without the spaces.

Second, the site is or used to be funny about certain words, triangle and angle being two of them. You need to use such words outside LaTeX before you use then in LaTeX.

Having used them once, I can then render \(\displaystyle \triangle\) or \(\displaystyle \angle\).

Third, is there a question in this post?
 
I edited the post.

My question is: How ratios are written here?
 
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Ganesh Ujwal said:
I edited the post.

My question is: How ratios are written here?
Click to expand...

I suppose you mean, "How are ratios written on this site?" [or in this problem, or in Latex, or something?]

Ratios are written just as you did: a:b, or a:b:c in your example.

Do you have a specific question about how ratios are written?
 
Dr.Peterson said:
I suppose you mean, "How are ratios written on this site?" [or in this problem, or in Latex, or something?]

Ratios are written just as you did: a:b, or a:b:c in your example.

Do you have a specific question about how ratios are written?
Click to expand...

Please forget about tex etc. I am referring to the problem.

I am asking to explain ratio step in the problem.
 
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Ganesh Ujwal said:
Please forget about tex etc. I am referring to the problem.

I am asking to explain ratio step in the problem.
Click to expand...

I am asking you to explain what you mean by "How ratios are written here?". Your English is not clear, so you need to say more.

It now appears that you mean something like "How is the ratio obtained at the step shown in red below?"

Until now, it has seemed that you had solved the problem and were just showing your (correct, though unnecessarily long) work. Evidently you actually copied it from some source, and you are asking for an explanation of it. Is that right? This should have been stated clearly.

Ganesh Ujwal said:
In the below figure, O is the center of the circle. Radius is 7 cm. \(\displaystyle \angle BOC\) = 120 Find the length of BC?

View attachment 9364

Solution: Join AO to meet BC in M

\(\displaystyle \Delta ABC\) \(\displaystyle \cong\) \(\displaystyle \Delta AOC\) and

\(\displaystyle \Delta BMO\) \(\displaystyle \cong\) \(\displaystyle \Delta CMO\)

\(\displaystyle \angle BMO\) = 90

Triangle BMO has three angles are 90, 60 and 30.

The sides are in the ratio of 2: \(\displaystyle \sqrt 3\) : 1

2X = 7 \(\displaystyle \Rightarrow\) X = 7/2

BM = \(\displaystyle \Rightarrow\) \(\displaystyle \sqrt 3\) x 7 / 2 = 7\(\displaystyle \sqrt 3 / 2\)

BC = 2 x 7\(\displaystyle \sqrt 3 / 2\) = 7 \(\displaystyle \sqrt 3\)
Click to expand...

If I am right, this is a well-known fact about the 30-60-90 triangle, because it is half of an equilateral triangle. The side opposite the right angle is twice as long as the leg opposite the 30 degree angle, and the other leg is sqrt(3) times that. This last fact is obtained from the Pythagorean theorem.

If this is not the answer to your question, please state your question as thoroughly as possible. What parts do you understand, and what do you not understand about the part you ask about?
 
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lev888 said:
"Triangle BMO has three angles are 90, 60 and 30."
Do these particular values tell you anything?
Click to expand...

Thank you for understanding my question.

Do these particular values tell you anything? - Yes, It gives me right angled triangle.
and also ratio values resembling trigonometric functions for special angles.
 
Ganesh Ujwal said:
In the below figure, O is the center of the circle. Radius is 7 cm. [ tex]\angle BOC[ /tex] = 120 Find the length of BC?

View attachment 9364

Solution: Join AO to meet BC in M

\(\displaystyle \Delta ABC\) \(\displaystyle \cong\) \(\displaystyle \Delta AOC\) and

\(\displaystyle \Delta BMO\) \(\displaystyle \cong\) \(\displaystyle \Delta CMO\)

\(\displaystyle \angle BMO\) = 90

Triangle BMO has three angles are 90, 60 and 30.

The sides are in the ratio of 2: \(\displaystyle \sqrt 3\) : 1

2X = 7 \(\displaystyle \Rightarrow\) X = 7/2

BM = \(\displaystyle \Rightarrow\) \(\displaystyle \sqrt 3\) x 7 / 2 = 7\(\displaystyle \sqrt 3 / 2\)

BC = 2 x 7\(\displaystyle \sqrt 3 / 2\) = 7 \(\displaystyle \sqrt 3\)
Click to expand...

if bmo is indeed 900 it is pretty straightforward,
but bc does not look perpendicular to oa to me.
if not, you could use law of sins (and/or cosines)?
 
How is the ratio obtained at the step shown in the solution? - Is my actual problem.

Forget about my diagram. Use your own diagram and explain to me.

My solution is right and answer came, but ratio part is confusing.
 
Ganesh Ujwal said:
Thank you for understanding my question.Do these particular values tell you anything? - Yes, It gives me right angled triangle.and also ratio values resembling trigonometric functions for special angles.
Click to expand...
What do 30 and 60 tell us? Have you had problems dealing with such triangles? E. g. calculate sides given one side and an angle?
 
Ganesh Ujwal said:
I got the answer from here: http://mathcentral.uregina.ca/qq/database/qq.09.05/gary1.html.

But how 2 X = 7 is obtained in the solution?
Click to expand...

I hope you recognize (perhaps from my hints) that you don't need to find the ratio by searching for a reference to it, but can derive it from the equilateral triangle. See, for example, http://www.themathpage.com/aTrig/30-60-90-triangle.htm#proof.

But in any case, the solution you are trying to understand is very poorly written (where did it come from??), so it is not much surprise that you are struggling with it! They never said what x stands for!

What they are saying is that, to find the distance OM from the center to the midpoint of the chord, you can recognize that the ratio of that distance (temporarily thought of as x) to the radius OB (given as 7) is 1:2, so we have x:7 = 1:2, which is equivalent to saying 2x = 7. I would just have said that the short leg of a 30-60-90 triangle is half as long as the hypotenuse, so this must be 7/2.
 
What they are saying is that, to find the distance OM from the center to the midpoint of the chord, you can recognize that the ratio of that distance (temporarily thought of as x) to the radius OB (given as 7) is 1:2. -> Which distance are you referring to?

You and I don't know other approach to the answer, So we have to follow that poorly written solution.

If you know other approach to find the length of BC. then post it.
 
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I got the answer myself: 2x = 7 because 2 x is hypotenuse. Here hypotenuse is radius of the circle i.e 7.
 
Ganesh Ujwal said:
What they are saying is that, to find the distance OM from the center to the midpoint of the chord, you can recognize that the ratio of that distance (temporarily thought of as x) to the radius OB (given as 7) is 1:2. -> Which distance are you referring to?

You and I don't know other approach to the answer, So we have to follow that poorly written solution.

If you know other approach to find the length of BC. then post it.
Click to expand...

"That distance" refers to the distance I just mentioned, OM. Yes, I should have just said OM to be clearer.

No, we don't have to follow a poorly written solution; we can rewrite it, or solve it ourselves. I presume there is some reason you want to understand that version; you haven't said where it comes from.

Yes, I do know another approach -- not very different, but details would be different if I wrote it. the important thing is not the approach (the content of the proof), but how well it communicates to the reader. Here is how I might have stated the problem and solution, trying to make both clear:

Problem: Two tangents AB and AC are constructed through point A to a circle with center O and radius 7. The measure of angle BOC is 120 degrees. Find the length of chord BC.

Solution: Construct the perpendicular to BC through O; this intersects BC at M. [It also passes through A, but this is not needed.] Then triangles BOM and COM are congruent right triangles, by the hypotenuse-leg theorem; so angle BOM = COM = 60 degrees, and BM = CM. Since BOM is a 30-60-90 triangle, the ratio of leg BM, opposite the 60 degree angle, to the hypotenuse OB, is sqrt(3)/2. It is given that OB, the radius, is 7 units long, so BM = sqrt(3)/2 * 7. The chord BC is twice this long, so BC = 7 sqrt(3).

Essentially the same idea, with a few minor differences. One thing that makes a proof well-written is that it explains the reasoning where needed for clarity; one has to decide how much to expect the reader to fill in.
 
Ganesh Ujwal said:
How is the ratio obtained at the step shown in the solution? - Is my actual problem.

Forget about my diagram. Use your own diagram and explain to me.

My solution is right and answer came, but ratio part is confusing.
Click to expand...

2 or 3 answers
1-the ratios for a 300, 600, 900 triangle are just known
2- or you just know the sin of 30=1/2, the cos=sqrt3/2
or/if you just remember the sin 300=1/2, then you have two sides, 1, and the hypotenuse=2, so you know the 3rd side has to be sqrt3, Pythagorean theorem.
 
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