Magnitudes 64704

The triangle ABC determines the size of the sides a and b and the magnitudes of the interior angles β and γ, given c = 1.86 m, the line on the side c is 2.12 m, and the angle alpha is 40 ° 12 '.

Correct answer:

a = 1.7865 m
b = 2.7436 m
β = 97.576 °
γ = 42.224 °

Step-by-step explanation:

c = 1.86 m m_{3} = 2.12 m α = 40 + 12 / 60 = \frac{2 0 1}{5} = 40.2^{\circ} m_{3}^{2} = (c / 2)^{2} + b^{2} - 2 (c / 2) b cos α k_{1} = cos (α) = cos (40.2 °) ≐ 0.7638 m_{3}^{2} = (c / 2)^{2} + b^{2} - 2 \cdot (c / 2) \cdot b \cdot k_{1} 2.1 2^{2} = (1.86 / 2)^{2} + b^{2} - 2 \cdot (1.86 / 2) \cdot b \cdot 0.76379602863464 - b^{2} + 1.421 b + 3.63 = 0 b^{2} - 1.421 b - 3.63 = 0 p = 1; q = - 1.421; r = - 3.63 D = q^{2} - 4 p r = 1.42 1^{2} - 4 \cdot 1 \cdot (- 3.63) = 16.5362765781 D > 0 b_{1, 2} = \frac{- q \pm D}{2 p} = \frac{1 . 4 2 \pm 1 6 . 5 4}{2} b_{1, 2} = 0.71033 \pm 2.033241 b_{1} = 2.743571351 b_{2} = - 1.322910738 b = b_{1} = 2.7436 ≐ 2.7436 a^{2} = b^{2} + c^{2} - 2 \cdot b \cdot c \cdot cos α a = b^{2} + c^{2} - 2 \cdot b \cdot c \cdot cos α = b^{2} + c^{2} - 2 \cdot b \cdot c \cdot cos 40.2 ° = 2.743 6^{2} + 1.8 6^{2} - 2 \cdot 2.7436 \cdot 1.86 \cdot cos 40.2 ° = 2.743 6^{2} + 1.8 6^{2} - 2 \cdot 2.7436 \cdot 1.86 \cdot 0.763796 = 1.786 = 1.7865 m

Our quadratic equation calculator calculates it.

β = \frac{1 8 0 °}{π} \cdot arccos (\frac{a ^{2} + c ^{2} - b ^{2}}{2 \cdot a \cdot c}) = \frac{1 8 0 °}{π} \cdot arccos (\frac{1 . 7 8 6 5 ^{2} + 1 . 8 6 ^{2} - 2 . 7 4 3 6 ^{2}}{2 \cdot 1 . 7 8 6 5 \cdot 1 . 8 6}) = 97.57 6^{\circ} = 97 ° 3 4^{'} 34 "

γ = 180 - α - β = 180 - 40.2 - 97.576 = 42.22 4^{\circ} = 42 ° 1 3^{'} 26 "

Try calculation via our triangle calculator.

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