Arusuc4eka2re
Arusuc4eka2re Arusuc4eka2re
  • 07-07-2016
  • Mathematics
contestada

two altitudes of an isosceles triangle are equal to 20 cm and 30 cm. Determine the possible measures of the base angles of the triangle

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Hagrid
Hagrid Hagrid
  • 14-07-2016
Given the image attached, where the altitude 30 is the one in red, 20 the one in green.

CE / AD = BC / AB
20 / 30 = 2 / 3

BC = 2 / 3 AB
BD = 1 / 2 BC = 1 / 3 AB

Using Pythgorean theorem
BD^2 + AD^2 = AB^2
(1/3AB)^2 + 30^2 = AB^2
(1/9AB)^2 + 900 = AB^2
AB^2 + 8100 = 9AB^2
8AB^2 = 8100
AB^2 = (8100/8)
AB = sqrt(8100/8)
AB = 45 / sqrt(2)

Looking for the base angle
sin B = AD/AB
sin B = 30 / [45/sqrt(2)]
sin B = 0.9426090416
B = sin^-1 (0.9426090416)
B = 70.53 degrees

Ver imagen Hagrid
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