In ΔOPQ, o = 850 cm, q = 800 cm and ∠Q=25°. Find all possible values of ∠O, to the nearest degree.

To solve this problem, we will use the Law of Cosines. The Law of Cosines states that for any triangle with sides of lengths a, b, and c and an angle γ opposite the side of length c, the following equation holds:

c² = a² + b² - 2ab cos(γ)

In this case, we have a = o = 850 cm, b = q = 800 cm, and γ = ∠Q = 25°. We want to find ∠O, so we need to rearrange the Law of Cosines to solve for this angle.

First, let's find the length of side p using the Law of Cosines:

p² = o² + q² - 2oq cos(∠Q) p² = (850 cm)² + (800 cm)² - 2(850 cm)(800 cm) cos(25°)

Calculate the above expression to find the value of p², and then take the square root to find the value of p.

Next, we can use the Law of Cosines again to find ∠O. This time, we have a = p, b = q = 800 cm, and c = o = 850 cm. The equation is:

o² = p² + q² - 2pq cos(∠O)

Rearrange this equation to solve for cos(∠O):

cos(∠O) = (p² + q² - o²) / (2pq)

Calculate the above expression to find the value of cos(∠O), and then use the inverse cosine function (also known as arccos) to find the value of ∠O. Remember that the result will be in radians, so you will need to convert it to degrees by multiplying by 180/π.

Finally, note that there are two possible values for ∠O: the one you found, and 180° minus that value. This is because the cosine function is positive in both the first and second quadrants. However, since ∠O is an angle in a triangle, it must be less than 180°, so only the smaller of the two possible values is valid.