Given cos A = 0.25 find angle A in degrees. Round your answer to the nearest hundredth.

1 answer:

8 0

We have that
cos A=0.25
so
A=arc cos (0.25)-------> using a calculator----> A=75.5225°
Round to the nearest hundredth-----> A=75.52²

the answer is
the option <span>75.52°</span>

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We choose the coordinates (4,4), (4,10), and (8,4) as the vertices of right-angle triangle ABC

We want to dilate triangle ABC by scale factor of 2 at the center (0,2)

The image is shown below as triangle A'B'C' with coordinates (8.6), (8,18), and (18,6).

Triangle ABC has a base of 4 units and a height of 6 units
Triangle A'B'C' has a base of 8 units and a height of 12 units

Applying Pythagoras theorem to triangle ABC to find the hypotenuse
$BC^{2}= AB^{2}+ AC^{2}$
$BC^{2}= 4^{2}+ 6^{2}$
$BC^{2} =52$
$BC=2 \sqrt{13}$

Now, we calculate the hypotenuse of the image
$B'C'^{2}= A'B'^{2}+ A'C'^{2}$
$B'C'^{2}= 8^{2}+ 12^{2}$
$B'C'^{2}=208$
$B'C'= \sqrt{208} =4 \sqrt{13}$

The hypotenuse of the image is twice the hypotenuse of triangle ABC which proof the two shapes are similar shape