# 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).

Result

A =  21.787 °
B =  38.213 °
C =  120 °

#### Solution:

$a = 3 \ \\ b = 5 \ \\ c = 7 \ \\ A_{ 1 } = \arccos((b^2+c^2-a^2)/(2 \cdot \ b \cdot \ c)) = \arccos((5^2+7^2-3^2)/(2 \cdot \ 5 \cdot \ 7)) \doteq 0.3803 \ \\ A = A_{ 1 } \rightarrow \ ^\circ = A_{ 1 } \cdot \ \dfrac{ 180 }{ \pi } \ \ ^\circ = 21.7867892983 \ \ ^\circ = 21.787 ^\circ = 21^\circ 47'12"$

Try calculation via our triangle calculator.

$B_{ 1 } = \arccos((a^2+c^2-b^2)/(2 \cdot \ a \cdot \ c)) = \arccos((3^2+7^2-5^2)/(2 \cdot \ 3 \cdot \ 7)) \doteq 0.6669 \ \\ B = B_{ 1 } \rightarrow \ ^\circ = B_{ 1 } \cdot \ \dfrac{ 180 }{ \pi } \ \ ^\circ = 38.2132107018 \ \ ^\circ = 38.213 ^\circ = 38^\circ 12'48"$
$C = 180-A-B = 180-21.7868-38.2132 = 120 = 120 ^\circ$

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