Points A, B, and C are on the circumference of a circle with radius 2 such that angle BAC = 45 degrees and angle ACB = 60 degrees. Find the area of triangle ABC.
In triangle ABC, the angles are
A=45
C=60
so, B=75
These are inscribed angles, so if you draw radii from the center O to each vertex, you have
AOB = 2*ACB = 120
BOC = 2*BAC = 90
AOC = 2*ABC = 150
Now you have three isosceles triangles. Each triangle with central angle θ has area
a = 1/4 r^2 sinθ
Now just plug and chug. If you really want punishment, use the sum-to product formulas to reduce the answer to a single product.
To find the area of triangle ABC, we can use the formula for the area of a triangle given two sides and the included angle. Let's call the center of the circle O.
Step 1: Find the length of side AB
Since points A and B are on the circumference of the circle centered at O with a radius of 2, the length of side AB is equal to the diameter of the circle, which is twice the radius:
AB = 2 * 2 = 4
Step 2: Find the length of side AC
Similarly, since points A and C are on the circumference of the circle centered at O with a radius of 2, the length of side AC is also equal to the diameter of the circle, which is twice the radius:
AC = 2 * 2 = 4
Step 3: Use the Law of Cosines to find the length of side BC
In triangle ABC, we can use the Law of Cosines to find the length of side BC. The Law of Cosines states that for any triangle with sides a, b, and c, and angle C (opposite to side c), the following equation holds:
c^2 = a^2 + b^2 - 2ab * cos(C)
In this case, side a = AB = 4, side b = AC = 4, and angle C = angle ACB = 60 degrees. Let's calculate BC:
BC^2 = 4^2 + 4^2 - 2 * 4 * 4 * cos(60)
BC^2 = 16 + 16 - 32 * 0.5
BC^2 = 32 - 16
BC^2 = 16
BC = sqrt(16)
BC = 4
Step 4: Calculate the area of triangle ABC
Now that we have the lengths of all three sides of triangle ABC, we can use the formula for the area of a triangle given two sides and the included angle:
Area = (1/2) * AB * AC * sin(angle BAC)
In this case, AB = 4, AC = 4, and angle BAC = 45 degrees. Let's calculate the area:
Area = (1/2) * 4 * 4 * sin(45)
Area = 8 * sqrt(2) / 2
Area = 4 * sqrt(2)
Therefore, the area of triangle ABC is 4 * sqrt(2) square units.
To find the area of triangle ABC, we can use the formula for the area of a triangle given two sides and the included angle. In this case, we know one side, which is the radius of the circle (2 units), and two angles (angle BAC and angle ACB).
Since we have the radius of the circle, triangle ABC is an isosceles triangle with AB = AC = 2 units.
To find the length of side BC, we can use the Law of Cosines, which states that in a triangle with sides a, b, and c and angle C opposite side c, the following equation holds:
c^2 = a^2 + b^2 - 2ab * cos(C)
In this case, a = 2 units (AB), b = 2 units (AC), and C = 60 degrees (angle ACB).
c^2 = 2^2 + 2^2 - 2(2)(2) * cos(60)
c^2 = 4 + 4 - 8 * (1/2)
c^2 = 8 - 4
c^2 = 4
c = 2
So, the length of side BC is also 2 units.
Now we know all three sides of triangle ABC (AB = AC = BC = 2 units).
To find the area, we can use the formula for the area of an isosceles triangle:
Area = (b^2 * sin(A))/2
In this case, b = 2 units (AB) and A = 45 degrees (angle BAC).
Area = (2^2 * sin(45))/2
Area = (4 * (sqrt(2)/2))/2
Area = (4 * sqrt(2)/2)/2
Area = 2 * (sqrt(2)/2)
Area = sqrt(2)
Therefore, the area of triangle ABC is sqrt(2) square units.