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!

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- Aug 30th 2006, 09:46 AM #1

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- Aug 30th 2006, 10:11 AM #2

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- Aug 30th 2006, 10:18 AM #3

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- Aug 30th 2006, 10:34 AM #4

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Originally Posted by**Neffets...:P**

(Look at the beautifully artistic picture).

---

You have by the Law of Sines,

$\displaystyle \frac{\sin 32^o}{4}=\frac{\sin C}{7}$

From here we have,

$\displaystyle \sin C=\frac{7}{4}\sin 32^o\approx .9274$

Then,

$\displaystyle C\approx 68.0268^o$

But that is not the only answer. There is one more in the second quadrant namely,

$\displaystyle 180^o-C\approx 111.9732^o$

- Aug 30th 2006, 10:38 AM #5

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- Aug 30th 2006, 10:46 AM #6

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Originally Posted by**Neffets...:P**

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, 10:50 AM #7

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- Aug 31st 2006, 08:04 AM #8

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- Aug 31st 2006, 01:47 PM #9

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Hello, Neffets!

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

How much can you increase $\displaystyle \angle A$, while $\displaystyle AB$ and $\displaystyle BC$ 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 $\displaystyle C = 90^o.$Code:B * * * 7 * * * * 4 * * * * A * * * * * * * * * * C

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