Triangle height angle

Calculate the lengths of the sides in an isosceles triangle, given the height (to the base) Vc= 8.8 cm and the angle at the base alpha= 38°40`.

Final Answer:

a =  14.0848 cm
b =  14.0848 cm
c =  21.9946 cm

Step-by-step explanation:

$$\begin{gathered} v_3=8.8\text{cm} \\ \alpha =38°40^{\prime }=38°+\frac{40}{60}°=38.6667°\doteq 38.6667 \\ \sin \alpha =v_3:a \\ a=v_3/\sin \alpha =v_3/\sin 38.666666666667°=8.8/\sin 38.666666666667°=8.8/0.624789=14.085=14.0848\text{cm} \end{gathered}$$

$$\begin{gathered} a=b \\ b=a=14.0848=14.0848\text{cm} \end{gathered}$$

$$\begin{gathered} \cos \alpha =(c/2):a \\ c/2=a\cdot \cos \alpha \\ c=2\cdot a\cdot \cos \alpha =2\cdot a\cdot \cos 38.666666666667°=2\cdot 14.0848\cdot \cos 38.666666666667°=2\cdot 14.0848\cdot 0.780794=21.995=21.9946\text{cm} \end{gathered}$$

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