130k views
1 vote
Help with geometry please‼️‼️

Help with geometry please‼️‼️-example-1
User Brugner
by
8.1k points

1 Answer

5 votes
To find the measure of angle A (m∠A) in triangle ABC, we can use the Law of Cosines, which states:

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

Where c is the side opposite angle C, and a and b are the lengths of the other two sides.

In this case, side AB (c) has a length of 32, side AC (a) has a length of 23, and side BC (b) has a length of 20. We want to find the measure of angle A (m∠A).

Plugging the values into the Law of Cosines equation, we have:

32^2 = 23^2 + 20^2 - 2 * 23 * 20 * cos(A)

Simplifying:

1024 = 529 + 400 - 920 * cos(A)

1024 = 929 - 920 * cos(A)

920 * cos(A) = 929 - 1024

920 * cos(A) = -95

cos(A) = -95 / 920

Now, to find the measure of angle A, we can take the inverse cosine (cos^-1) of (-95 / 920):

m∠A = cos^-1(-95 / 920)

Using a calculator, we find:

m∠A ≈ 96.0 degrees

Therefore, the measure of angle A (m∠A) is approximately 96.0 degrees.
User Oogles
by
8.0k points

No related questions found