In triangle ABC, measure of angle A is 30 degrees. AC=4, AB=3 square root of 3. Find BC.
clear case of the cosine law
BC^2 = 4^2 + (3√3)^2 - 2(4)(3√3)cos30°
carry on
square root of 7?
To find the length of BC in triangle ABC, we can use the law of cosines. The law of cosines states that in any triangle, the square of one side length is equal to the sum of the squares of the other two sides, minus twice the product of the other two sides multiplied by the cosine of the included angle.
In this case, we know the lengths of sides AC and AB, as well as the measure of angle A. Let's label BC as c, AC as a, and AB as b.
Using the law of cosines, we have:
c^2 = a^2 + b^2 - 2ab*cos(A)
Plugging in the values we know:
c^2 = 4^2 + (3√3)^2 - 2*4*(3√3)*cos(30°)
Simplifying further:
c^2 = 16 + 27 - 24√3*cos(30°)
cos(30°) = √3/2, so let's substitute that value:
c^2 = 16 + 27 - 24√3*(√3/2)
simplifying:
c^2 = 16 + 27 - 36
c^2 = 7
Taking the square root of both sides, we get:
c = √7
So the length of BC is √7 units.