Right triangle

Calculate the length of the remaining two sides and the angles in the rectangular triangle ABC if a = 10 cm, angle alpha = 18°40'.

Correct result:

A =  18.6667 °
B =  71.3333 °
C =  90 °
a =  10 cm
b =  29.6004 cm
c =  31.244 cm

Solution:

$A=18+40\mathrm{/}60={\displaystyle \frac{56}{3}}=18\frac{2}{3}=18.6667^{\circ }=18^{\circ }40^{\mathrm{\prime }}$

$B=90-A=90-18.6667={\displaystyle \frac{214}{3}}=71\frac{1}{3}=71.3333^{\circ }=71^{\circ }20^{\mathrm{\prime }}$

$C=90=90^{\circ }$

$a=10=10\text{ cm}$

$$\begin{gathered} \tan A=a\mathrm{/}b \\ b=a\mathrm{/}\tan (A^{\circ }\to \text{ rad})=a\mathrm{/}\tan (A^{\circ }\cdot {\displaystyle \frac{\pi }{180}})=10\mathrm{/}\tan (18.6666666667^{\circ }\cdot {\displaystyle \frac{3.1415926}{180}})=29.6=29.6004\text{ cm} \end{gathered}$$

$c=\sqrt{a^2+b^2}=\sqrt{10^2+29.6004^2}=31.244\text{ cm}$

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