vector please.....

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January 27th 2007, 10:12 PM
sallsa

vector please.....

If I have the triangle ABC, that A(3,4) B(1,2) C(6,1)
How can I find the value of angle A? :confused:
Please, thank you
January 27th 2007, 11:16 PM
earboth

Quote:

Originally Posted by sallsa

If I have the triangle ABC, that A(3,4) B(1,2) C(6,1)
How can I find the value of angle A? :confused:
Please, thank you

Hello,

from the title of your post I assume that you have to use vectors (If you don't want to use vectors you can solve this problem by using Cosine rule)

1. Calculation of the vectors:

$\overrightarrow{AB}=(1,2)-(3,4)=(-2,-2)$

$\overrightarrow{AC}=(6,1)-(3,4)=(3,-3)$

2. The angle $\alpha$ between 2 vectors $\vec u$ and $\vec v$ is calculated by:

$\cos(\alpha)=\frac{\vec u \cdot \vec v}{|\vec u| \cdot |\vec v|}$

Plug in the vectors you know into this formula:

$\cos(A)=\frac{(-2,-2) \cdot (3,-3)}{\sqrt{8} \cdot \sqrt{18}}=\frac{0}{12}=0$

Therefore A = 90° : You are dealing with a right triangle.

3. Hint: Draw a sketch of this triangle to prove the calculations.

EB
January 27th 2007, 11:17 PM
CaptainBlack

Quote:

Originally Posted by sallsa

If I have the triangle ABC, that A(3,4) B(1,2) C(6,1)
How can I find the value of angle A? :confused:
Please, thank you

Calculate the squares of the lengths of the sides:

AB^2=(3-1)^2+(4-2)^2=8

AC^2=(3-6)^2+(4-1)^2=18

BC^2=(1-6)^2+(2-1)^2=26.

So AB^2+AC^2=BC^2. Therefore by the converse of Pythagoras's theorem
ABC is a right triangle with BC as hypotenuse, and the angle at A is a right
angle.

RonL