In a triangle ABC AC = BC, angle C is 120 degrees, AC = 25√3. Find AB

Given: △ ABC, AC = BC, ∠С = 120 °, AC = 25 Sqrt3 cm.
Find: AB.
If sides AC and BC are equal, then △ ABC is isosceles. Draw the height CH from the top of C to the bottom of AB.
Since ∠С = 120 °, then 60 ° falls on the cuts A and B, based on the theorem on the sum of the angles of a triangle.
Based on the properties of an isosceles triangle, the angles at the base of AB are equal. Means:
∠А = ∠В = 30 °.
△ ASN – rectangular, since CH – height.
For the property of a right-angled triangle, CH = 1/2 AC (leg opposite to an angle of 30 °). We have:
CH = 1/2 * 25√3 = 12.5√3 (cm).
Now you can find the AH leg. Behind the Pythagorean theorem:
AC ^ 2 = CH ^ 2 + AH ^ 2.
Hence:
AH ^ 2 = AC ^ 2 – CH ^ 2 = (25√3) ^ 2 – (12.5√3) ^ 2 = (625 * 3) – (156.25 * 3) = 1875 – 468.75 = 1406.25 (cm).
AH = Sqrt1406.25 = 37.5 (cm).
For the property of an isosceles triangle, the height of CH △ ABC is also its bisector and median.
For the median property:
AB = 37.5 * 2 = 75 (cm).
Answer: AB = 75 cm.




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