# Triangle TBC

TBC is isosceles triangle with base TB with base angle 63° and legs length |TC| = |BC| = 25. How long is the base TB?

Result

|TB| =  22.7

#### Solution:

$\cos 63 ^\circ = \dfrac{|TB|/2}{ |TC|} \ \\ \ \\ |TB| = 2 \cdot 25 \cdot \cos(63 ^\circ ) = 22.7$

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