Let's go through the problems step by step.

Problem 9: Find PQ.

We have a right triangle with the given angle of 85° and the side QR = 9. To find PQ, we can use trigonometric ratios. Since we have an angle and the side opposite to PQ, we use the tangent function:

$\tan(85^\circ) = \frac{\text{opposite}}{\text{adjacent}} = \frac{PQ}{QR}$

Substitute the known values:

$\tan(85^\circ) = \frac{PQ}{9}$

Solve for PQ:

$PQ = 9 \times \tan(85^\circ)$

Problem 10: Find DE.

In this right triangle, we know the length of side DF = 13, and the angle D = 21°. To find DE, we use the sine or cosine function. Since we are given the adjacent and need to find the hypotenuse, we use the cosine:

$\cos(21^\circ) = \frac{DE}{13}$

Solve for DE:

$DE = 13 \times \cos(21^\circ)$

Problem 11: Find ∠A.

We have a right triangle with sides 3.0 and 6.8. To find angle A, we use the tangent function:

$\tan(A) = \frac{\text{opposite}}{\text{adjacent}} = \frac{6.8}{3.0}$

Take the inverse tangent:

$A = \tan^{-1}\left(\frac{6.8}{3.0}\right)$

Problem 12: Find ∠R.

This triangle has sides 24 and 9. To find angle R, again use the tangent function:

$\tan(R) = \frac{24}{9}$

Find R:

$R = \tan^{-1}\left(\frac{24}{9}\right)$

Problem 13: Find ∠B.

For the right triangle with sides 17 and 16, use the sine or cosine function. Since we have the adjacent side and want to find angle B, use the tangent function:

$\tan(B) = \frac{16}{17}$

Then, use the inverse tangent:

$B = \tan^{-1}\left(\frac{16}{17}\right)$

Would you like me to calculate the numerical values for these problems, or is there any particular part you need more explanation on?