In an isosceles triangle ABC AB = BC = 4, angle B = 120 degrees, M and N are the midpoints of AB

| March 10, 2021 | education

In an isosceles triangle ABC AB = BC = 4, angle B = 120 degrees, M and N are the midpoints of AB and BC, respectively. Find: dot product BA * AC MN * AC.

In the triangle ABC, by the cosine theorem, we define the length of the AC side.

AC ^ 2 = BA ^ 2 + BC ^ 2 – 2 * BA * BC * Cos120 = 16 + 16 – 2 * 4 * 4 * (-1 / 2) = 32 + 16 = 48 cm.

AC = 4 * √3 cm.

To determine the dot product of the vectors BA and AC, it is necessary to combine the beginning of the vectors at one point.

Let’s move the beginning of the AC vector to point B, for which we will make a parallel translation of the vector.

Then the angle АА1С1 = 120 + 30 = 1500.

Let’s define the scalar product of vectors BA and AC.

VA * AC = | VA | * | AC | * Cos150 = 4 * 4 * √3 * (-√3 / 2) = -24.

Since points M and H are the middle of the lateral sides, then MH is the middle line of the triangle ABC, then MH = AC / 2 = 4 * √3 / 2 = 2 * √3 cm, and MH is parallel to AC.

The vectors МН and АС coincide in direction, then the angle between them is equal to 00.

Let’s define the scalar product of vectors МН and АС.

МН * АС = | МН | * | AC | * Cos0 = 2 * √3 * 4 * √3 * 1 = 24.

Answer: The dot product of the vectors is -24, 24.



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