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In right triangle [tex]\(ABC\)[/tex], [tex]\(\angle A\)[/tex] and [tex]\(\angle B\)[/tex] are complementary angles and [tex]\(\sin A=\frac{8}{9}\)[/tex]. What is [tex]\(\cos B\)[/tex]?

A. [tex]\(\frac{8 \sqrt{17}}{17}\)[/tex]

B. [tex]\(\frac{8}{9}\)[/tex]

C. [tex]\(\frac{\sqrt{17}}{9}\)[/tex]

D. [tex]\(\frac{\sqrt{17}}{8}\)[/tex]

Sagot :

GinnyAnswer
Certainly! Let's solve the given problem step-by-step.

In a right triangle [tex]\( \triangle ABC \)[/tex] with [tex]\( \angle C \)[/tex] being the right angle, the angles [tex]\( \angle A \)[/tex] and [tex]\( \angle B \)[/tex] are complementary. This means that [tex]\( \angle A + \angle B = 90^\circ \)[/tex].

Given:
[tex]\[ \sin A = \frac{8}{9} \][/tex]

We need to find [tex]\( \cos B \)[/tex].

1. Complementary angles in a right triangle have a special relationship with the sine and cosine functions:
[tex]\[ \sin A = \cos B \][/tex]

2. Therefore, since [tex]\( \angle A \)[/tex] and [tex]\( \angle B \)[/tex] are complementary angles:
[tex]\[ \cos B = \sin A \][/tex]

3. Substitute the given value:
[tex]\[ \cos B = \frac{8}{9} \][/tex]

Thus, the value of [tex]\( \cos B \)[/tex] is [tex]\( \frac{8}{9} \)[/tex].

Therefore, the correct choice is:
[tex]\[ \boxed{\frac{8}{9}} \][/tex]
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