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7 practice 7 (from unit 4, lesson 10) here is triangle abc. select all …

Question

7 practice 7 (from unit 4, lesson 10)
here is triangle abc.
select all expressions which give the measure of angle a.
a (arccosleft(\frac{28}{53}
ight))
b (arccosleft(\frac{45}{53}
ight))
c (arcsinleft(\frac{28}{53}
ight))
d (arcsinleft(\frac{45}{53}
ight))
e (arctanleft(\frac{28}{75}
ight))
f (arctanleft(\frac{45}{28}
ight))

Explanation:

To find the measure of angle \( A \) in right triangle \( ABC \) (right - angled at \( C \)):

Step 1: Recall trigonometric ratios

In a right - triangle, for an acute angle \( \theta \):

  • \( \cos\theta=\frac{\text{adjacent}}{\text{hypotenuse}} \)
  • \( \sin\theta=\frac{\text{opposite}}{\text{hypotenuse}} \)
  • \( \tan\theta=\frac{\text{opposite}}{\text{adjacent}} \)

For angle \( A \):

  • The adjacent side to angle \( A \) is \( AC = 28 \)
  • The opposite side to angle \( A \) is \( BC=45 \)
  • The hypotenuse is \( AB = 53 \)
Step 2: Analyze option A (\( \arccos(\frac{28}{53}) \))

Using the cosine ratio \( \cos A=\frac{\text{adjacent to }A}{\text{hypotenuse}}=\frac{AC}{AB}=\frac{28}{53} \). If \( \cos A = \frac{28}{53} \), then \( A=\arccos(\frac{28}{53}) \). So option A is correct.

Step 3: Analyze option B (\( \arccos(\frac{45}{53}) \))

\( \cos A=\frac{28}{53}
eq\frac{45}{53} \), so \( A
eq\arccos(\frac{45}{53}) \). Option B is incorrect.

Step 4: Analyze option C (\( \arcsin(\frac{28}{53}) \))

\( \sin A=\frac{\text{opposite to }A}{\text{hypotenuse}}=\frac{BC}{AB}=\frac{45}{53}
eq\frac{28}{53} \), so \( A
eq\arcsin(\frac{28}{53}) \). Option C is incorrect.

Step 5: Analyze option D (\( \arcsin(\frac{45}{53}) \))

Since \( \sin A=\frac{BC}{AB}=\frac{45}{53} \), then \( A = \arcsin(\frac{45}{53}) \). Option D is correct.

Step 6: Analyze option E (\( \arctan(\frac{28}{45}) \))

\( \tan A=\frac{\text{opposite to }A}{\text{adjacent to }A}=\frac{BC}{AC}=\frac{45}{28}
eq\frac{28}{45} \), so \( A
eq\arctan(\frac{28}{45}) \). Option E is incorrect.

Step 7: Analyze option F (\( \arctan(\frac{45}{28}) \))

Since \( \tan A=\frac{\text{opposite to }A}{\text{adjacent to }A}=\frac{BC}{AC}=\frac{45}{28} \), then \( A=\arctan(\frac{45}{28}) \). Option F is correct.

Answer:

A. \( \arccos(\frac{28}{53}) \), D. \( \arcsin(\frac{45}{53}) \), F. \( \arctan(\frac{45}{28}) \)