# How much?

• Aug 30th 2006, 10:46 AM
Neffets...:P
How much?
Hi...

In a triangle ABC are AB= 7 cm, BC= 4cm and angle BAC = 32 degrees...

How much can you increase angle BAC, while AB and BC are unchanged?

Can you guys help me on this?

Would be appreciated! :)
• Aug 30th 2006, 11:11 AM
ThePerfectHacker
Quote:

Originally Posted by Neffets...:P

How much can you increase angle BAC, while AB and BC are unchanged?

I really do not understand. Do you mean find the side the opposite side?
• Aug 30th 2006, 11:18 AM
Neffets...:P
...
No, I'm not trying to find the opposite side...

Kinda hard to explain this... How large can the angle BAC be if AB and BC have the length i wrote before? In other words increase the angle BAC to the point where CB "falls" off AC. Understand?

:confused:
• Aug 30th 2006, 11:34 AM
ThePerfectHacker
Quote:

Originally Posted by Neffets...:P
. Understand?

Yes. So you are working with the "ambigous case".
(Look at the beautifully artistic picture).
---
You have by the Law of Sines,
$\frac{\sin 32^o}{4}=\frac{\sin C}{7}$
From here we have,
$\sin C=\frac{7}{4}\sin 32^o\approx .9274$
Then,
$C\approx 68.0268^o$
But that is not the only answer. There is one more in the second quadrant namely,
$180^o-C\approx 111.9732^o$
• Aug 30th 2006, 11:38 AM
Neffets...:P
• Aug 30th 2006, 11:46 AM
CaptainBlack
Quote:

Originally Posted by Neffets...:P
Hi...

In a triangle ABC are AB= 7 cm, BC= 4cm and angle BAC = 32 degrees...

How much can you increase angle BAC, while AB and BC are unchanged?

Can you guys help me on this?

Would be appreciated! :)

Consider C moving on a circle of radius 4cm around B. Every position is a valid
triangle with AB=7 and BC=4cm, and all such triangles occur for some
position of C on this circle.

The largest value of angle BAC occurs when AC is a tangent to this
circle, and angle ACB is a right angle. Then sin(BAC)=4/7, or the maximum
angle BAC=34.85 degrees. (You might want to try to prove this)

RonL

RonL
• Aug 30th 2006, 11:50 AM
CaptainBlack
Quote:

Originally Posted by ThePerfectHacker
Yes. So you are working with the "ambigous case".
(Look at the beautifully artistic picture).
---
You have by the Law of Sines,
$\frac{\sin 32^o}{4}=\frac{\sin C}{7}$
From here we have,
$\sin C=\frac{7}{4}\sin 32^o\approx .9274$
Then,
$C\approx 68.0268^o$
But that is not the only answer. There is one more in the second quadrant namely,
$180^o-C\approx 111.9732^o$

You have still not understood the question. In the base case angle BAC is
32 degrees. What is the largest size of this angle for a triangle with AB and
BC maintaining the given values.

RonL
• Aug 31st 2006, 09:04 AM
Neffets...:P
..
Thats right Captain!

You got it right...

Thanks for all of your help :D
• Aug 31st 2006, 02:47 PM
Soroban
Hello, Neffets!

Quote:

In $\Delta ABC,\;AB= 7\text{ cm, }BC= 4\text{ cm and }\angle A = 32^o$

How much can you increase $\angle A$, while $AB$ and $BC$ are unchanged?

From the given information, there are two possible triangles.
Code:

                        B                          B                         *                          *                     * *                        *  *               7  *  *                    7  *      *               *    * 4                  *          * 4             *      *                  *              *         * 32°    *                * 32°              *     A * * * * * * * C          A * * * * * * * * * * * * * C

The limit is when $C = 90^o.$
Code:

                        B                         *                     *  *               7  *    *               *        * 4             *          *         *              *     A * * * * * * * * * * C

Then $A \:=\:\sin^{-1}\left(\frac{4}{7}\right) \:\approx\:34.85^o$