Triangle tangent area

In the triangle ABC, b=5 cm, c=6 cm, /BAC/ = 80° are given. Calculate the sizes of the other sides and angles, and further determine the sizes of the tangent tc and the area of the triangle.

Final Answer:

a =  7.112 cm
β =  43.8166 °
γ =  56.1834 °
t =  5.3657 cm
S =  14.7721 cm²

Step-by-step explanation:

$$\begin{gathered} b=5\text{cm} \\ c=6\text{cm} \\ \alpha =80{}^{\circ } \\ {a}^2=b^2+c^2-2\cdot b\cdot c\cdot \cos \alpha \\ a=\sqrt{b^2+c^2-2\cdot b\cdot c\cdot \cos \alpha }=\sqrt{b^2+c^2-2\cdot b\cdot c\cdot \cos 80°}=\sqrt{5^2+6^2-2\cdot 5\cdot 6\cdot \cos 80°}=\sqrt{5^2+6^2-2\cdot 5\cdot 6\cdot 0.173648}=7.112=7.112\text{cm} \end{gathered}$$

$$\begin{gathered} b^2=a^2+c^2-2\cdot a\cdot c\cdot \cos \beta \\ B=\arccos (\frac{a^2+c^2-b^2}{2\cdot a\cdot c})=\arccos (\frac{7.112^2+6^2-5^2}{2\cdot 7.112\cdot 6})\doteq 0.7647\text{rad} \\ \beta =B\to \text{°}=B\cdot \frac{180}{\pi }\text{°}=0.7647\cdot \frac{180}{\pi }\text{°}=43.817\text{°}=43.8166^{\circ }=43°49^{\prime } \end{gathered}$$

$\gamma =180-\alpha -\beta =180-80-43.8166=56.1834^{\circ }=56°11^{\prime }$

Try calculation via our triangle calculator.

$$\begin{gathered} t^2=b^2+(c/2{)}^2-2\cdot b\cdot (c/2)\cdot \cos \alpha ) \\ t=\sqrt{b^2+(c/2{)}^2-b\cdot c\cdot \cos \alpha }=\sqrt{b^2+(c/2{)}^2-b\cdot c\cdot \cos 80°}=\sqrt{5^2+(6/2{)}^2-5\cdot 6\cdot \cos 80°}=\sqrt{5^2+(6/2{)}^2-5\cdot 6\cdot 0.173648}=5.366=5.3657\text{cm} \end{gathered}$$

$$\begin{gathered} s=\frac{a+b+c}{2}=\frac{7.112+5+6}{2}\doteq 9.056\text{cm} \\ S=\sqrt{s\cdot (s-a)\cdot (s-b)\cdot (s-c)}=\sqrt{9.056\cdot (9.056-7.112)\cdot (9.056-5)\cdot (9.056-6)}=14.7721{\text{cm}}^2 \end{gathered}$$

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