Angles by cosine law

Calculate the size of the angles of the triangle ABC, if it is given by: a = 3 cm; b = 5 cm; c = 7 cm (use the sine and cosine theorem).

Correct result:

A =  21.7868 °
B =  38.2132 °
C =  120 °

Solution:

$$\begin{gathered} a=3 \\ b=5 \\ c=7 \\ {A}_1=\arccos ((b^2+c^2-a^2)\mathrm{/}(2\cdot b\cdot c))=\arccos ((5^2+7^2-3^2)\mathrm{/}(2\cdot 5\cdot 7))\doteq 0.3803 \\ A=A_1{\to }^{\circ }=A_1\cdot {{\displaystyle \frac{180}{\pi }}}^{\circ }=0.3803\cdot {{\displaystyle \frac{180}{\pi }}}^{\circ }=21.787^{\circ }=21.7868^{\circ }=21^{\circ }47^{\mathrm{\prime }}12\mathrm{"} \end{gathered}$$

Try calculation via our triangle calculator.

$$\begin{gathered} B_1=\arccos ((a^2+c^2-b^2)\mathrm{/}(2\cdot a\cdot c))=\arccos ((3^2+7^2-5^2)\mathrm{/}(2\cdot 3\cdot 7))\doteq 0.6669 \\ B=B_1{\to }^{\circ }=B_1\cdot {{\displaystyle \frac{180}{\pi }}}^{\circ }=0.6669\cdot {{\displaystyle \frac{180}{\pi }}}^{\circ }=38.213^{\circ }=38.2132^{\circ }=38^{\circ }12^{\mathrm{\prime }}48\mathrm{"} \end{gathered}$$

$C=180-A-B=180-21.7868-38.2132=120^{\circ }$

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