Using sin^2(α) + cos^2(α) = 1, we find cos(α) = √(1 - 0.6^2) = √(1 - 0.36) = √0.64 = 0.8.

The angles where sin(θ) = 0 in the interval [0, 2π] are θ = 0 and θ = π.

sin(θ) = 0 at θ = 0° and 180°.

Using the identity sin^2(θ) + cos^2(θ) = 1, we have cos^2(θ) = 1 - (1/√2)^2 = 1 - 1/2 = 1/2. Thus, cos(θ) = ±1/√2.

sin(θ) = 1/√2 at θ = 45°.

sin(θ) = 1/√2 at θ = 45° and θ = 225°.

sin(θ) = 1/√2 at θ = 45°.

Using the Pythagorean identity, cos(θ) = √(1 - sin²(θ)) = √(1 - (3/5)²) = 4/5.

Using the identity tan(θ) = sin(θ)/cos(θ) and cos(θ) = √(1 - sin^2(θ)), we find cos(θ) = 3/5. Thus, tan(θ) = (4/5)/(3/5) = 4/3.

Using the identity tan A = sin A / cos A, we can find sin A = tan A * cos A. Using the Pythagorean identity, we find sin A = 3/5.

Since tan θ = sin θ / cos θ and tan θ = 1, we have sin θ = cos θ. For θ = 45°, sin θ = 1/√2.

If tan(x) = 1, then sin(x) = cos(x). Therefore, sin(x) + cos(x) = 2sin(x) = 2(1/√2) = √2.

tan(45°) = 1, hence x = 45°.

Using the identity tan(x) = sin(x)/cos(x), we can find sin(x) = 3/5 after applying the Pythagorean theorem.

Using the identity tan(x) = sin(x)/cos(x), we can find sin(x) = 3/5 after calculating the hypotenuse using the Pythagorean theorem.

tan(θ) = 1 at θ = 45°.

Using the identity sin²(θ) + cos²(θ) = 1, we find sin(θ) = 3/5.

Height = shadow * tan(angle) = 10 * √3 = 5√3 m.

Let the angles be 2x, 3x, and 4x. Then, 2x + 3x + 4x = 180 degrees. Thus, 9x = 180 degrees, x = 20 degrees. The largest angle is 4x = 80 degrees.

Let the angles be 2x, 3x, and 4x. Then, 2x + 3x + 4x = 180. So, 9x = 180, x = 20. The largest angle = 4x = 80 degrees.

Area = 1/2 * base * height. Therefore, 30 = 1/2 * 10 * height, height = 6 cm.

Area = 1/2 * base * height => 30 = 1/2 * 10 * height => height = 30 / 5 = 6.

Area = 1/2 * base * height. Thus, 30 = 1/2 * 10 * height. Height = 6.

Area = (1/2) * base * height. Therefore, 60 = (1/2) * 12 * height, height = 10 cm.

Area = 1/2 * base * height. 30 = 1/2 * 10 * height. Therefore, height = 6 cm.

Using the formula R = (abc)/(4 * Area), we can find the side opposite to angle A. Let a = side opposite to A. Then, a = (4 * Area * R) / (bc) = (4 * 48 * 10) / (b * c).

The maximum area of a triangle with circumradius R is given by the formula Area = (abc)/(4R). For maximum area, the triangle should be equilateral, thus Area = (3√3/4) * (R^2) = (3√3/4) * (5^2) = 25√3/4 cm².

The area of a triangle is given by (1/2) * base * height = (1/2) * 10 * 12 = 60 cm².

Area = (4/3) * √[m1 * m2 * m3] = (4/3) * √[6 * 8 * 10] = 48 cm².

Area = (4/3) * √(s(s - m1)(s - m2)(s - m3)), where s = (6 + 8 + 10)/2 = 12. Area = (4/3) * √(12 * 6 * 4 * 2) = 48.