Difference between revisions of "2030 AMC 8 Problems/Problem 3"
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== Solution== | == Solution== | ||
| − | Triangle ABC is a right triangle with its right angle at B. Therefore, 𝐴𝐶 is the hypotenuse of right triangle ABC, and 𝐴𝐵 and 𝐵𝐶 are the legs of right triangle ABC. According to the Pythagorean | + | Triangle ABC is a right triangle with its right angle at B. Therefore, 𝐴𝐶 is the hypotenuse of right triangle ABC, and 𝐴𝐵 and 𝐵𝐶 are the legs of right triangle ABC. According to the Pythagorean Thereom: |
| − | + | 𝐴𝐵 = √20^2 − 162^2 = √400 − 256 = 144 = 12 | |
Since triangle DEF is similar to triangle ABC, with vertex F corresponding to vertex C, the measure of angle ∠𝐹 equals the measure of angle∠𝐶. Therefore, 𝑠𝑖𝑛𝐹 = 𝑠𝑖𝑛𝐶. From the side lengths of triangle ABC, | Since triangle DEF is similar to triangle ABC, with vertex F corresponding to vertex C, the measure of angle ∠𝐹 equals the measure of angle∠𝐶. Therefore, 𝑠𝑖𝑛𝐹 = 𝑠𝑖𝑛𝐶. From the side lengths of triangle ABC, | ||
| − | + | sinF = opposite side/hypotenuse = 𝐴𝐵/𝐴𝐶 = 12/20 = 3/5 | |
Therefore, 𝑠𝑖𝑛𝐹 = 3/5. | Therefore, 𝑠𝑖𝑛𝐹 = 3/5. | ||
Revision as of 21:13, 23 June 2024
Problem
In triangle 𝐴𝐵𝐶, the measure of ∠𝐵 is 90°, 𝐵𝐶=16, and 𝐴𝐶=20. Triangle 𝐷𝐸𝐹 is similar to triangle 𝐴𝐵𝐶, where vertices 𝐷, 𝐸, and 𝐹 correspond to vertices 𝐴, 𝐵, and 𝐶, respectively, and each side of triangle 𝐷𝐸𝐹 is 13 the length of the corresponding side of triangle 𝐴𝐵𝐶. What is the value of 𝑠𝑖𝑛𝐹?
Solution
Triangle ABC is a right triangle with its right angle at B. Therefore, 𝐴𝐶 is the hypotenuse of right triangle ABC, and 𝐴𝐵 and 𝐵𝐶 are the legs of right triangle ABC. According to the Pythagorean Thereom: 𝐴𝐵 = √20^2 − 162^2 = √400 − 256 = 144 = 12
Since triangle DEF is similar to triangle ABC, with vertex F corresponding to vertex C, the measure of angle ∠𝐹 equals the measure of angle∠𝐶. Therefore, 𝑠𝑖𝑛𝐹 = 𝑠𝑖𝑛𝐶. From the side lengths of triangle ABC, sinF = opposite side/hypotenuse = 𝐴𝐵/𝐴𝐶 = 12/20 = 3/5 Therefore, 𝑠𝑖𝑛𝐹 = 3/5.
See also
| 2030 AMC 8 (Problems • Answer Key • Resources) | ||
| Preceded by Problem 2 |
Followed by Problem 4 | |
| 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | ||
| All AJHSME/AMC 8 Problems and Solutions | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
