Given \sin A=\frac{5}{\sqrt{89}}sinA=
89
​

5
​
and that angle AA is in Quadrant I, find the exact value of \cos AcosA in simplest radical form using a rational denominator.

Answer:

First don't forget the [ tex ] [ /tex ] in order to execute the latex code.

Step-by-step explanation:

[tex] \sin A=\frac{5}{\sqrt{89}}sinA= 89 [/tex]

[tex] \cos AcosA [/tex] = [tex] \cos [/tex]

The exact value of cosine of angle A for the considered case using a rational denominator in simplest radical form is 8√(89)/89

Trigonometric ratios for a right angled triangle are from the perspective of a particular non-right angle.

In a right angled triangle, two such angles are there which are not right angled(not of 90 degrees).

The slant side is called hypotenuse.

From the considered angle, the side opposite to it is called perpendicular, and the remaining side will be called base.

From that angle (suppose its measure is θ),

[tex]\sin(\theta) = \dfrac{\text{Length of perpendicular}}{\text{Length of Hypotenuse}}\\\cos(\theta) = \dfrac{\text{Length of Base }}{\text{Length of Hypotenuse}}\\\\\tan(\theta) = \dfrac{\text{Length of perpendicular}}{\text{Length of base}}\\\\\cot(\theta) = \dfrac{\text{Length of base}}{\text{Length of perpendicular}}\\\\\sec(\theta) = \dfrac{\text{Length of Hypotenuse}}{\text{Length of base}}\\\\\csc(\theta) = \dfrac{\text{Length of Hypotenuse}}{\text{Length of perpendicular}}\\[/tex]

If ABC is a triangle with AC as the hypotenuse and angle B with 90 degrees then we have:

[tex]|AC|^2 = |AB|^2 + |BC|^2[/tex]

where |AB| = length of line segment AB. (AB and BC are rest of the two sides of that triangle ABC, AC being the hypotenuse).

Given that:

[tex]\sin A=\dfrac{5}{\sqrt{89}}[/tex]

So, from this, we are given the ratio of perpendicular and hypotenuse from the perspective of angle A.

There must be existing a right triangle with same measure of angle A such that:

Perpendicular's length = P = 5 units

and Hypotenuse' length = H = [tex]\sqrt{89}[/tex] units.

Using Pythagoras theorem, we get:

Base' length = B as:

[tex]P^2 + B^2 = H^2\\B = \sqrt{H^2 - P^2}\\\\B = \sqrt{(\sqrt{89})^2 - 5^2}\\\\B = \sqrt{89-25} = \sqrt{64} = 8[/tex]

(took positive root as B is representing length, a non-negative quantity).

Thus, Base is of 8 units length.

Thus, we get:

[tex]\cos(\theta) = \dfrac{\text{Length of Base }}{\text{Length of Hypotenuse}}\\\\\cos(A) = \dfrac{B}{H} = \dfrac{8}{\sqrt{89}}[/tex]

Rationalizing, we get:

[tex]\cos(A) = \dfrac{8}{\sqrt{89}} =\dfrac{8}{\sqrt{89}} \times \dfrac{\sqrt{89}}{\sqrt{89}} \\\\\cos(A) = \dfrac{8\sqrt{89}}{89}[/tex]

Thus, the exact value of cosine of angle A for the considered case using a rational denominator in simplest radical form is [tex]\dfrac{8\sqrt{89}}{89}[/tex]

Learn more about Pythagoras theorem here:

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