# Triangle (Proving)

• Jan 11th 2010, 03:33 AM
cloud5
Triangle (Proving)
I'm getting very confused...there are too many formula for this! (Headbang)

Prove:
1. sin (A+B) + cos (A-B) $\displaystyle \equiv$ (sin A+ cos A)(sin B + cos B)

2. tan (A+B) - tan A $\displaystyle \equiv \frac{sin B}{cos A cos (A+B)}$

And one more extra:
3. In triangle ABC, AB=p, BC=3p, and $\displaystyle \angle ABC=60^o$. The bisector of the angle ABC meets AC at the point T. Find AT and CT in terms of p.

Answer: $\displaystyle AT=\frac{p}{4}\sqrt{7}; CT=\frac{3p}{4}\sqrt{7}$
• Jan 11th 2010, 04:30 AM
Quote:

Originally Posted by cloud5
I'm getting very confused...there are too many formula for this! (Headbang)

Prove:
1. sin (A+B) + cos (A-B) $\displaystyle \equiv$ (sin A+ cos A)(sin B + cos B)

2. tan (A+B) - tan A $\displaystyle \equiv \frac{sin B}{cos A cos (A+B)}$

And one more extra:
3. In triangle ABC, AB=p, BC=3p, and $\displaystyle \angle ABC=60^o$. The bisector of the angle ABC meets AC at the point T. Find AT and CT in terms of p.

Answer: $\displaystyle AT=\frac{p}{4}\sqrt{7}; CT=\frac{3p}{4}\sqrt{7}$

HI

(1) $\displaystyle \sin A\cos B+\cos A\sin B+\cos A\cos B+\sin A\sin B$

$\displaystyle \sin A(\cos B+\sin B)+\cos A(\cos B+\sin B)$

$\displaystyle (\sin A+\cos A)(\sin B+\cos B)$

(2) $\displaystyle \frac{\sin (A+B)}{\cos (A+B)}-\frac{\sin A}{\cos A}$

$\displaystyle \frac{\sin (A+B)\cos A-\sin A\cos (A+B)}{\cos A\cos (A+B)}$

$\displaystyle \frac{\sin A\cos B\cos A+\cos A\sin B\cos A-\cos A\cos B\sin A+\sin^2 A\sin B}{\cos A\cos (A+B)}$

$\displaystyle \frac{(\cos^2 A+\sin^2 A)\sin B}{\cos A\cos (A+B)}$

so the result follow .

(3) Make a sketch first . The ratio of AT:TC=1:3

By the cosine rule , $\displaystyle AC^2=p^2+(3p)^2-2p(3p)\cos 60$

$\displaystyle AC=p\sqrt{7}$

$\displaystyle AT=\frac{1}{4}AC=\frac{1}{4}p\sqrt{7}$

sure you can find TC now .
• Jan 12th 2010, 03:20 AM
cloud5
Quote:

Originally Posted by mathaddict
$\displaystyle \frac{\sin A\cos B\cos A+\cos A\sin B\cos A-\cos A\cos B\sin A+\sin^2 A\sin B}{\cos A\cos (A+B)}$

$\displaystyle \frac{(\cos^2 A+\sin^2 A)\sin B}{\cos A\cos (A+B)}$

so the result follow .

Can you explain what happen between those steps?
• Jan 12th 2010, 03:26 AM
Quote:

Originally Posted by cloud5
Can you explain what happen between those steps?

sure , notice that \cos A\cos B\sin A cancels off and you are left with

$\displaystyle \frac{\cos^2 A\sin B+\sin^2 A\sin B}{\cos A\cos (A+B)}$

then bring out the common factor , sin B .
• Jan 12th 2010, 03:40 AM
cloud5
Quote:

Originally Posted by mathaddict

$\displaystyle AC=p\sqrt{7}$

$\displaystyle AT=\frac{1}{4}AC=\frac{1}{4}p\sqrt{7}$

sure you can find TC now .

I don't understand this part... how you got$\displaystyle \sqrt{7}$?
• Jan 12th 2010, 03:49 AM
Quote:

Originally Posted by cloud5
I don't understand this part... how you got$\displaystyle \sqrt{7}$?

Note that cos 60=1/2

$\displaystyle AC^2=10p^2-6p^2(\frac{1}{2})$

$\displaystyle AC^2=7p^2 so AC=p\sqrt{7}$
• Jan 14th 2010, 12:57 AM
cloud5
Quote:

Originally Posted by mathaddict

$\displaystyle AT=\frac{1}{4}AC=\frac{1}{4}p\sqrt{7}$

One more question, how you get $\displaystyle \frac{1}{4}$?
• Jan 14th 2010, 02:29 AM
Quote:

Originally Posted by cloud5
One more question, how you get $\displaystyle \frac{1}{4}$?

because AT:TC is in the ratio of 1:3 , the 'total' for AC is 4 .

so AT = 1/4 AC
• Jan 14th 2010, 03:17 AM
cloud5
Quote:

Originally Posted by mathaddict
because AT:TC is in the ratio of 1:3 , the 'total' for AC is 4 .

so AT = 1/4 AC

How to see the ratio(it dosen't state in the question)? Is it from the sketch? Sorry I know I'm asking too much! (Crying)