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How do i go about solving this question? - straight line - HSN forum

# How do i go about solving this question? - straight line

3 replies to this topic

### #1umhellothere

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Posted 15 September 2008 - 05:38 PM

The right-angled triangle OAB with sides of length 3cm, 4cm and 5cm is placed with one vertex at the origin as shown in the diagram.

A circle centre C and diameter RO of length 13cm is drawn and passes through O and B.

What is the gradient of the line RO?

My scanner isn't working so i've attached a (really bad) sketch of what the diagram looks like, i didn't draw the circle because my paint skills ain't up to that much!

*I have been trying for ages to get this answer but my mind just goes blank everytime i look at it. I don't have any idea of what formula/method to use so just a nudge in the right direction would help me a lot thanks for any help you can give*

### #2Steve

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Posted 16 September 2008 - 09:03 AM

Welcome to the forum!

I'll give you a little hint: it relies on the fact from Standard Grade/Int2 that a triangle in a semi-circle is right-angled. You can then use trigonometry to work out the angle between OR and the x-axis.

Hope this helps.
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### #3umhellothere

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Posted 16 September 2008 - 03:51 PM

QUOTE(Steve @ Sep 16 2008, 10:03 AM)
Welcome to the forum!

I'll give you a little hint: it relies on the fact from Standard Grade/Int2 that a triangle in a semi-circle is right-angled. You can then use trigonometry to work out the angle between OR and the x-axis.

Hope this helps.

yes it really helped! Does -3.93 sound about right for the answer? , thanks for getting me started

### #4Steve

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Posted 18 September 2008 - 08:19 AM

That's not what I got, but I could be wrong!

Here's my working:

Since the triangle in the semi-circle is right-angled, RB = 12 (using Pythagoras's Theorem). Then using trigonometry, angle ROB = 78.91 degrees and angle BOA = 36.87 degrees.

So OR makes an angle of 78.91 + 36.87 = 115.78 degrees with the x-axis. Then we can use the formula to find the gradient. I get -2.07.
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