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 =  21.787 °
B =  38.213 °
C =  120 °

Solution:

a=3 b=5 c=7 A1=arccos⁡((b2+c2−a2)/(2⋅ b⋅ c))=arccos⁡((52+72−32)/(2⋅ 5⋅ 7))≐0.3803 A=A1→ ∘=A1⋅ 180π  ∘=0.38025120669293⋅ 180π  ∘=21.787  ∘=21.787∘=21∘47′12"a=3 \ \\ b=5 \ \\ c=7 \ \\ A_{1}=\arccos((b^2+c^2-a^2)/(2 \cdot \ b \cdot \ c))=\arccos((5^2+7^2-3^2)/(2 \cdot \ 5 \cdot \ 7)) \doteq 0.3803 \ \\ A=A_{1} \rightarrow \ ^\circ =A_{1} \cdot \ \dfrac{ 180 }{ \pi } \ \ ^\circ =0.38025120669293 \cdot \ \dfrac{ 180 }{ \pi } \ \ ^\circ =21.787 \ \ ^\circ =21.787 ^\circ =21^\circ 47'12"

Try calculation via our triangle calculator.

B1=arccos⁡((a2+c2−b2)/(2⋅ a⋅ c))=arccos⁡((32+72−52)/(2⋅ 3⋅ 7))≐0.6669 B=B1→ ∘=B1⋅ 180π  ∘=0.66694634450366⋅ 180π  ∘=38.213  ∘=38.213∘=38∘12′48"B_{1}=\arccos((a^2+c^2-b^2)/(2 \cdot \ a \cdot \ c))=\arccos((3^2+7^2-5^2)/(2 \cdot \ 3 \cdot \ 7)) \doteq 0.6669 \ \\ B=B_{1} \rightarrow \ ^\circ =B_{1} \cdot \ \dfrac{ 180 }{ \pi } \ \ ^\circ =0.66694634450366 \cdot \ \dfrac{ 180 }{ \pi } \ \ ^\circ =38.213 \ \ ^\circ =38.213 ^\circ =38^\circ 12'48"
C=180−A−B=180−21.7868−38.2132=120∘C=180-A-B=180-21.7868-38.2132=120 ^\circ

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Tips to related online calculators
Cosine rule uses trigonometric SAS triangle calculator.
See also our trigonometric triangle calculator.
Pythagorean theorem is the base for the right triangle calculator.

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