Isosceles triangle

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

Final Answer:

A =  44.9009 °
B =  44.9009 °
C =  90.1982 °
c =  24.0832 cm

Step-by-step explanation:

$$\begin{gathered} r=17\text{cm} \\ h=12\text{cm} \\ \sin A=h:r \\ A=\frac{180°}{\pi }\cdot \arcsin (h/r)=\frac{180°}{\pi }\cdot \arcsin (12/17)=44.9009^{\circ }=44°54^{\prime }3" \end{gathered}$$

$B=A=44.9009=44.9009^{\circ }=44°54^{\prime }3"$

$C=180-A-B=180-44.9009-44.9009=90.1982^{\circ }=90°11^{\prime }54"$

$$\begin{gathered} (c/2{)}^2+h^2=r^2 \\ c=2\cdot \sqrt{r^2-h^2}=2\cdot \sqrt{17^2-12^2}=2\sqrt{145}=24.0832\text{cm} \end{gathered}$$

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