Question

AB=2cm, AC=3cm, BAC=60' BC=??

Answer 1

. По теореме косинусов:

BC² = AC² + AB² - 2AC•AB•cosBAC

BC² = 4 + 9 - 2•2•3•1/2 = 13 - 6 = 7 => BC = √7.

2. По теореме косинусов:

BC² = AC² + AB² - 2AC•AB•cosА

Откуда cosA = (AB² + AC² - BC²)/(2•AB•AC)

cosA =(49 + 36 - 25)/(7•6•2) = 60/ 84 = 15/ 21 = 5/7

Answer 2

To find the length of side BC in a triangle ABC where AB = 2 cm, AC = 3 cm, and angle BAC = 60 degrees, you can use the law of cosines. The law of cosines relates the lengths of the sides of a triangle to the cosine of one of its angles.

The law of cosines is given by:

c^2 = a^2 + b^2 - 2ab * cos(C)

In this case:

• c is the side opposite angle BAC, which is BC.
• a is the side opposite angle ACB, which is AB.
• b is the side opposite angle ABC, which is AC.
• C is the angle BAC, which is 60 degrees.

So, you can plug in the values:

BC^2 = (2 cm)^2 + (3 cm)^2 - 2 * 2 cm * 3 cm * cos(60°)

Now, calculate the values inside the equation:

BC^2 = 4 cm^2 + 9 cm^2 - 12 cm^2 * cos(60°)

Now, calculate cos(60°):

cos(60°) = 0.5

Now, substitute this value back into the equation:

BC^2 = 4 cm^2 + 9 cm^2 - 12 cm^2 * 0.5

BC^2 = 4 cm^2 + 9 cm^2 - 6 cm^2

BC^2 = 13 cm^2 - 6 cm^2

BC^2 = 7 cm^2

Now, take the square root of both sides to find BC:

BC = √(7 cm^2)

BC ≈ √7 cm

BC ≈ 2.65 cm (rounded to two decimal places)

So, the length of side BC is approximately 2.65 cm.