Cosine

Cosine and sine theorem:
Calculate all unknown values (sides and angles) of the triangle ABC.

a = 20 cm; b = 15 cm; γ = 90°; c =? cm; α =? °; β =? °

Correct answer:

c =  25 cm
α =  53.1301 °
β =  36.8699 °

Step-by-step explanation:

$$\begin{gathered} a=20\text{cm} \\ b=15\text{cm} \\ \gamma =90{}^{\circ } \\ {c}^2=a^2+b^2-2\cdot a\cdot b\cdot \cos \gamma \\ c=\sqrt{a^2+b^2-2\cdot a\cdot b\cdot \cos \gamma }=\sqrt{a^2+b^2-2\cdot a\cdot b\cdot \cos 90°}=\sqrt{20^2+15^2-2\cdot 20\cdot 15\cdot \cos 90°}=\sqrt{20^2+15^2-2\cdot 20\cdot 15\cdot 0}=25=25\text{cm} \\ \text{ Verifying Solution: } \\ {c}_2=\sqrt{a^2+b^2}=\sqrt{20^2+15^2}=25\text{cm} \\ \cos 90°=\cos \pi /2=0 \\ \sin 90°=\sin \pi /2=1 \end{gathered}$$

$$\begin{gathered} \sin \alpha :\sin \gamma =a:c \\ \alpha =\frac{180°}{\pi }\cdot \arcsin (\frac{a}{c}\cdot \sin (\pi /2))=\frac{180°}{\pi }\cdot \arcsin (\frac{20}{25}\cdot \sin (3.1416/2))=53.1301^{\circ }=53°7^{\prime }48" \end{gathered}$$

$$\begin{gathered} \sin \beta :\sin \gamma =b:c \\ \beta =\frac{180°}{\pi }\cdot \arcsin (\frac{b}{c}\cdot \sin (\pi /2))=\frac{180°}{\pi }\cdot \arcsin (\frac{15}{25}\cdot \sin (3.1416/2))=36.8699^{\circ }=36°52^{\prime }12" \end{gathered}$$

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