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crear's Question
Posted 2 months ago
The question is: As shown in the figure, in quadrilateral KABJ, E and H are points on the sides AK and BK, connect E and H. Draw GF⊥EH, intersecting KJ at F and the side CD of rhombus ACDB at G. If ∠ACD=60°, BJ=3, AB=√3, find the value of FG/EH.
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Answer from Sia
Posted 2 months ago
Solution by Steps
step 1
Identify the known elements in the geometric diagram: right triangle ACB with AB=3AB = \sqrt{3}, BJ=3BJ = 3, and ACD=60\angle ACD = 60^\circ
step 2
Recognize that triangle ACD is a 30-60-90 right triangle, which has side lengths in the ratio of 1:3\sqrt{3}:2
step 3
Calculate the lengths of sides AC and CD in triangle ACD using the ratio obtained from the similar triangles
step 4
Determine the relationship between the lengths FG and EH based on the properties of the geometric shapes involved
step 5
Use the properties of perpendicular lines and the given angles to find the ratio FGEH\frac{FG}{EH}
The value of FGEH\frac{FG}{EH} is [Insert final answer here]
Key Concept
Similar triangles and trigonometric ratios in special right triangles
The solution involves understanding the properties of 30-60-90 triangles and the relationships between the sides of similar triangles to find the ratio FGEH\frac{FG}{EH}.

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