Isosceles triangle 10

In an isosceles triangle, the equal sides are 2/3 of the length of the base. Determine the measure of the base angles.

Result

B =  41.41 °

Solution:

a=c=23b cosB=b/2a cosB=b/223b cosB=1/223  B1=arccos(1/22/3)0.7227 rad  B=B1 =B1 180π  =41.4096221092  =41.41=412435"a = c = \dfrac{ 2 }{ 3 } b \ \\ \cos B = \dfrac{ b/2 }{ a } \ \\ \cos B = \dfrac{ b/2 }{ \dfrac{ 2 }{ 3 } b } \ \\ \cos B = \dfrac{ 1/2 }{ \dfrac{ 2 }{ 3 } } \ \\ \ \\ B_{ 1 } = \arccos(\dfrac{ 1/2 }{ 2/3 } ) \doteq 0.7227 \ rad \ \\ \ \\ B = B_{ 1 } \rightarrow \ ^\circ = B_{ 1 } \cdot \ \dfrac{ 180 }{ \pi } \ \ ^\circ = 41.4096221092 \ \ ^\circ = 41.41 ^\circ = 41^\circ 24'35"

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