Difference between revisions of "2030 AMC 8 Problems/Problem 3"

(Solution)
(Solution)
 
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𝐴𝐵 = √20^2 − 162^2 = √400 − 256 = 144 = 12
 
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
+
sinF = opposite side/hypotenuse = 𝐴𝐵/𝐴𝐶 = 12/20 = 3/5.
 
Therefore, 𝑠𝑖𝑛𝐹 = 3/5.
 
Therefore, 𝑠𝑖𝑛𝐹 = 3/5.
  

Latest revision as of 21:14, 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 (ProblemsAnswer KeyResources)
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

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