# Isosceles triangle

Calculate the size of the interior angles and the length of the base of the isosceles triangle if the length of the arm is 17 cm and the height to the base is 12 cm.

Correct result:

A =  44.901 °
B =  44.901 °
C =  90.198 °
c =  24.083 cm

#### Solution:

$r=17 \ \text{cm} \ \\ h=12 \ \text{cm} \ \\ \ \\ \sin A=h:r \ \\ \ \\ A=\dfrac{ 180^\circ }{ \pi } \cdot \arcsin (h/r)=\dfrac{ 180^\circ }{ \pi } \cdot \arcsin (12/17)=44.901 ^\circ =44^\circ 54'3"$
$B=A=44.9009=44.901 ^\circ =44^\circ 54'4"$
$C=180-A-B=180-44.9009-44.901=90.198 ^\circ =90^\circ 11'53"$
$(c/2)^2+h^2=r^2 \ \\ \ \\ c=2 \cdot \ \sqrt{ r^2-h^2 }=2 \cdot \ \sqrt{ 17^2-12^2 }=2 \ \sqrt{ 145 }=24.083 \ \text{cm}$

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