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17. In right triangle Delta ABC shown below, the given lengths are in millimeters .What is sinA A. (4sqrt (2))/(9) B. (4sqrt (2))/(7) C. (7sqrt (2))/(8) D. . (7)/(9) E. . (9)/(7)

Question

17. In right triangle Delta ABC shown below, the given lengths
are in millimeters .What is sinA
A. (4sqrt (2))/(9)
B. (4sqrt (2))/(7)
C. (7sqrt (2))/(8)
D. . (7)/(9)
E. . (9)/(7)

17. In right triangle Delta ABC shown below, the given lengths are in millimeters .What is sinA A. (4sqrt (2))/(9) B. (4sqrt (2))/(7) C. (7sqrt (2))/(8) D. . (7)/(9) E. . (9)/(7)

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UriahProfessional · Tutor for 6 years

Answer

To find \(\sin A\) in right triangle \(\triangle ABC\), we need to identify the lengths of the sides relative to angle A. In a right triangle, the sine of an angle is the ratio of the length of the opposite side to the length of the hypotenuse.Given:- AB = \(4\sqrt{2}\) mm (opposite side to angle A)- BC = 7 mm (adjacent side to angle A)- AC = 9 mm (hypotenuse)Since \(\sin A\) is the ratio of the length of the side opposite to angle A to the length of the hypotenuse, we have:\[\sin A = \frac{AB}{AC}\]Now we substitute the given lengths:\[\sin A = \frac{4\sqrt{2}}{9}\]Therefore, the correct answer is:A. \(\frac{4 \sqrt{2}}{9}\)
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