2 isoceles triangle have equl vetical angles and their areas are in ratio16:25 find the ratio of their corresponding heights
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Isosceles triangle: Base be b other two sides = a top angle = Ф height = h
Then other two angles will be = (180 - Ф) /2 = 90 - Ф/2
sin 90 - Ф/2 = h / a => cos Ф/2 = h / a => h = a cos Ф/2
a² = h² + b²/4 => b² = 4 (a² - h²) => b = 4 a sin Ф/2
Area = 1/2 a h = 1/2 a b cos Ф/2 = 1/2 a 4 a sin Ф/2 cos Ф/2
Area = 2 a² sin Ф/2 cos Ф/2 or can be written also as 1/2 h² / cos Ф/2
A1 : A2 = 16 : 25 = (a1)² / (a2)² = (h1)² / (h2)²
Ratio of sides = ratio of sq root of areas = 4 : 5 = ratio of heights
Then other two angles will be = (180 - Ф) /2 = 90 - Ф/2
sin 90 - Ф/2 = h / a => cos Ф/2 = h / a => h = a cos Ф/2
a² = h² + b²/4 => b² = 4 (a² - h²) => b = 4 a sin Ф/2
Area = 1/2 a h = 1/2 a b cos Ф/2 = 1/2 a 4 a sin Ф/2 cos Ф/2
Area = 2 a² sin Ф/2 cos Ф/2 or can be written also as 1/2 h² / cos Ф/2
A1 : A2 = 16 : 25 = (a1)² / (a2)² = (h1)² / (h2)²
Ratio of sides = ratio of sq root of areas = 4 : 5 = ratio of heights
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