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Equations, derivatives, integrals, matrices, triangles, primes, statistics — or a word problem the tutor breaks into parts.

Triangle 10 10 with angle 120

10,\ 10

Қадами ба қадам

  1. a = 10,\ b = 10,\ C = 120^\circ

    Two sides and the included angle (SAS): find the third side with the law of cosines.

  2. c^2 = a^2 + b^2 - 2ab\cos C = 300

    Law of cosines.

  3. c = 10 \sqrt{3} \approx 17.320

  4. P = a + b + c = 10 \sqrt{3} + 20

    Perimeter.

  5. s = \tfrac{P}{2} = 5 \sqrt{3} + 10,\quad A = \sqrt{s(s-a)(s-b)(s-c)} = 25 \sqrt{3} \approx 43.301

    Heron's formula for the area.

  6. \cos A = \frac{b^2 + c^2 - a^2}{2bc} = \frac{\sqrt{3}}{2} \Rightarrow A \approx 30.000^\circ

    Law of cosines for angle A (opposite side a).

  7. \cos B = \frac{a^2 + c^2 - b^2}{2ac} = \frac{\sqrt{3}}{2} \Rightarrow B \approx 30.000^\circ

    Law of cosines for angle B (opposite side b).

  8. \cos C = \frac{a^2 + b^2 - c^2}{2ab} = - \frac{1}{2} \Rightarrow C \approx 120.00^\circ

    Law of cosines for angle C (opposite side c).

  9. A + B + C = 180.0^\circ

    The angles add to 180° — a obtuse, isosceles triangle.

Ҷавоби ҷавобро нишон диҳед
A = 25 \sqrt{3} \approx 43.301,\quad P = 10 \sqrt{3} + 20,\quad \angle \approx 30.0^\circ, 30.0^\circ, 120.0^\circ