Question

Given the vertices of the triangle A (1; 1), B (4; 1), C (4; 5). Find the cosine of the angle A of this triangle.​

Answer 1

Answer:

Using law of cosines, we have, a

2

=b

2

+c

2

−2bccosA

From the given points, a

2

=BC

2

=68, b

2

=AC

2

=169, c

2

=AB

2

=25

Substituting these values, we have cosA=

2bc

b

2

+c

2

−a

2

cosA=

2×13×5

169+25−68

=

130

126

=

65

63

Answer 2

Answer:

Cos A = 4

Step-by-step explanation:

a = AB

b = AC

c = AB

FOR Cosine A, we have a law of Cosine And the formula is below,

$\boxed{\sf a² = b² + c² - 2bc \: \cos \: A}$

------------------------------

For a²,

BC² = a² Hence,

BC² = $\sf (c_1 - b_1)² + (c_2 + b_2)²$

BC² = $\sf (4 - 4)² + (5 - 1)²$

BC² = $\sf (0)² + (4)²$

BC² = $\sf 16$

------------------------------

For b²,

AC² = b² Hence,

AC² = $\sf (c_1 - a_1 )² + (c_2 - a_2)²$

AC² = $\sf (4-1)² + (5-1)²$

AC² = $\sf (3)² + (4)²$

AC² = $\sf 9 + 16$

AC² = $\sf 25$

------------------------------

For c²,

AB² = c² Hence,

AB² = $\sf (b_1 - a_1)² + (b_2 - a_2)²$

AB² = $\sf (4 - 1)² + (1 - 1)²$

AB² = $\sf (3)² + (0)²$

AB² = $\sf 9$

------------------------------

And also,

a² = 16 = BC²

b² = 25 = AC²

c² = 9 = AB²

-----------------

Therefore,

→ a² = 16

→ a = √16

→ a = 4

-----------------

→ b² = 25

→ b = 5

-----------------

→ c² = 9

→ c = 3

-----------------

Now, We have all values of Formula so we need to put it on,

$\sf a² = b² + c² - 2bc \: \cos \: A$

$\sf 16 = 25 + 9 - 2(5)(3)\: \cos \: A$

$\sf 16 = 34 - 30 \: \cos \: A$

$\sf 16 = 4 \: \cos \: A$

$\sf \cos \: A = \frac{16}{4}$

$\sf \cos \: A = \frac{\cancel{16}^{\: \: \cancel4×4}}{\cancel4}$

$\bf \underline{\underline{ \cos \: A = 4}}$

I hope it helps you :)