Problem 17: Find the measure of angle A in triangle ABC.

Let’s solve problem 17 first:
We are given triangle ABC with point D on AB. We know:

- AC = 35
- CD = 25
- DB = 12
- Angle CDB = 118°
We need to find angle A (m∠A) to the nearest whole degree.

Step 1: Look at triangle CDB.
We know two sides and the included angle? Wait — actually, we know side CD = 25, side DB = 12, and angle at D is 118°. That’s SAS (side-angle-side), so we can use the Law of Cosines to find side CB.

Law of Cosines:
CB² = CD² + DB² - 2·CD·DB·cos(angle CDB)

But wait — angle CDB is 118°, which is obtuse. Cosine of 118° is negative, so the last term becomes positive.

Calculate:

CB² = 25² + 12² - 2·25·12·cos(118°)

First, compute squares:
25² = 625
12² = 144
Sum = 625 + 144 = 769

Now cos(118°). Let’s calculate that.
cos(118°) ≈ cos(180° - 62°) = -cos(62°) ≈ -0.4695

So:

-2·25·12·(-0.4695) = +2·25·12·0.4695
= 600 · 0.4695 ≈ 281.7

So CB² ≈ 769 + 281.7 = 1050.7
Then CB ≈ √1050.7 ≈ 32.41

Okay, now we have CB ≈ 32.41

Step 2: Now look at triangle ACD or triangle ABC? Actually, we want angle A, which is in triangle ACD or ABC. But we don’t know AD yet.

Wait — maybe better to consider triangle ACD and triangle CDB together.

Actually, let’s try using Law of Sines in triangle CDB first to find angle DCB or angle CBD.

In triangle CDB:

We have sides: CD = 25, DB = 12, CB ≈ 32.41, angle at D = 118°

Use Law of Sines to find angle at B (angle CBD):

sin(angle CBD) / CD = sin(angle CDB) / CB

So:

sin(angle CBD) / 25 = sin(118°) / 32.41

sin(118°) = sin(180° - 62°) = sin(62°) ≈ 0.8829

So:

sin(angle CBD) = 25 * (0.8829 / 32.41) ≈ 25 * 0.02724 ≈ 0.681

Wait, that can’t be right — 0.8829 / 32.41 ≈ 0.02724? No!

Wait: 0.8829 ÷ 32.41 ≈ ?

Let me recalculate:

0.8829 / 32.41 ≈ 0.02724? That’s way too small.

No! 32.41 × 0.027 = about 0.875, yes — but then 25 × 0.02724 = 0.681 — that’s possible.

sin(angle CBD) ≈ 0.681 → angle CBD ≈ arcsin(0.681) ≈ 43°

Check: sin(43°) ≈ 0.682 — yes, close enough.

So angle CBD ≈ 43°

Now, in triangle CDB, angles sum to 180°:

Angle at D = 118°, angle at B ≈ 43°, so angle at C (angle DCB) = 180 - 118 - 43 = 19°

Okay, so angle DCB ≈ 19°

Now, look at triangle ACD. We know:

AC = 35, CD = 25, and we need angle at A.

But we don’t know any angles in triangle ACD yet. However, note that angle ACB = angle ACD + angle DCB.

But we don’t know angle ACD.

Alternative approach: Use Law of Cosines in triangle ACD? But we don’t know AD.

Wait — perhaps we can find length AB or AD?

Another idea: Use coordinates.

Place point D at origin (0,0). Since angle CDB = 118°, and DB = 12, place point B at (12, 0).

Point C: from D, distance 25, at an angle of 180° - 118° = 62° above the negative x-axis? Wait.

If D is at (0,0), and B is at (12,0), and angle at D is 118°, that means the angle between vectors DC and DB is 118°.

Vector DB is along positive x-axis: (12,0)

So vector DC makes 118° with DB. So if DB is along 0°, then DC is at 118° from it — so direction of DC is 118° from positive x-axis.

So coordinates of C:

x = 25 * cos(118°)
y = 25 * sin(118°)

cos(118°) ≈ -0.4695
sin(118°) ≈ 0.8829

So C_x = 25 * (-0.4695) ≈ -11.7375
C_y = 25 * 0.8829 ≈ 22.0725

So C is at approximately (-11.74, 22.07)

Now, point A is somewhere on the line extending from D through... wait, D is on AB, and A-D-B are colinear, with D between A and B? The diagram shows A--D--B, so yes.

Since D is at (0,0), and B is at (12,0), then A is on the x-axis to the left of D, say at (-p, 0) for some p > 0.

We know AC = 35. Point A is at (-p, 0), point C is at (-11.74, 22.07)

Distance AC = 35:

So:

√[ (-p - (-11.74))² + (0 - 22.07)² ] = 35

Simplify:

√[ (-p + 11.74)² + (-22.07)² ] = 35

Square both sides:

(-p + 11.74)² + (22.07)² = 35²

Compute:

22.07² ≈ 487.0849
35² = 1225

So:

(-p + 11.74)² = 1225 - 487.0849 ≈ 737.9151

Take square root:

-p + 11.74 ≈ ±√737.9151 ≈ ±27.165

So two cases:

Case 1: -p + 11.74 = 27.165 → -p = 15.425 → p = -15.425 → not possible since p>0

Case 2: -p + 11.74 = -27.165 → -p = -38.905 → p = 38.905

So point A is at (-38.905, 0)

Now, we need angle at A, which is angle between vectors AC and AB.

Point A: (-38.905, 0)
Point C: (-11.74, 22.07)
Point B: (12, 0)

Vector AC = C - A = (-11.74 - (-38.905), 22.07 - 0) = (27.165, 22.07)
Vector AB = B - A = (12 - (-38.905), 0 - 0) = (50.905, 0)

Angle at A is between vectors AC and AB.

Use dot product formula:

cos(theta) = (AC • AB) / (|AC| |AB|)

AC • AB = (27.165)(50.905) + (22.07)(0) ≈ 27.165 * 50.905

Calculate: 27.165 * 50.905 ≈ let's compute:

27 * 50.905 ≈ 1374.435
0.165 * 50.905 ≈ 8.40
Total ≈ 1382.835

|AC| = 35 (given)
|AB| = distance from A to B = 50.905 (since from x=-38.905 to x=12, difference 50.905)

So cos(theta) = 1382.835 / (35 * 50.905) ≈ 1382.835 / 1781.675 ≈ 0.7762

Then theta = arccos(0.7762) ≈ ?

cos(39°) ≈ 0.7771 — very close!

So angle A ≈ 39°

Let me verify calculation:

35 * 50.905 = 35 * 50 + 35 * 0.905 = 1750 + 31.675 = 1781.675 — correct.

Dot product: 27.165 * 50.905

Let me compute more accurately:

27.165 * 50 = 1358.25
27.165 * 0.905 = let's see: 27.165 * 0.9 = 24.4485, 27.165 * 0.005 = 0.135825, total 24.584325
So total dot product = 1358.25 + 24.584325 = 1382.834325 — same as before.

1382.834325 / 1781.675 ≈ ?

Divide: 1382.834325 ÷ 1781.675 ≈ 0.7762

arccos(0.7762): using calculator, cos(39°) = 0.7771, cos(39.1°) = ?

Difference: 0.7771 - 0.7762 = 0.0009

Derivative of cos is -sin, sin(39°)≈0.6293, so dtheta ≈ -d(cos)/sin ≈ 0.0009 / 0.6293 ≈ 0.00143 radians ≈ 0.082 degrees

So angle ≈ 39° + 0.082° ≈ 39.08° — still rounds to 39°

To be precise, let's use direct computation.

We can also use Law of Cosines in triangle ABC now that we have all sides? But we don't have BC exactly, but earlier we had approx 32.41.

From coordinates:

A(-38.905,0), B(12,0), C(-11.74,22.07)

AB = 50.905
AC = 35
BC = distance from B to C: dx = -11.74 - 12 = -23.74, dy = 22.07 - 0 = 22.07
BC² = (-23.74)^2 + (22.07)^2 = 563.5876 + 487.0849 = 1050.6725 → BC ≈ 32.415 — matches earlier.

Now in triangle ABC, sides:

AB = c = 50.905
BC = a = 32.415
AC = b = 35

Angle at A is opposite side BC? No.

Standard notation: angle at A is between sides AB and AC, so opposite side is BC.

Law of Cosines:

cos(A) = (b² + c² - a²) / (2bc)

Where b = AC = 35, c = AB = 50.905, a = BC = 32.415

So:

cos(A) = (35² + 50.905² - 32.415²) / (2 * 35 * 50.905)

Compute each:

35² = 1225
50.905² ≈ 2591.319025
32.415² ≈ 1050.732225

Numerator: 1225 + 2591.319025 - 1050.732225 = (1225 - 1050.732225) + 2591.319025 = 174.267775 + 2591.319025 = 2765.5868

Denominator: 2 * 35 * 50.905 = 70 * 50.905 = 3563.35

So cos(A) = 2765.5868 / 3563.35 ≈ 0.7762 — same as before.

arccos(0.7762) = ?

Using calculator: arccos(0.7762) ≈ 39.08 degrees.

So to nearest whole degree: 39°

Now, just to make sure, let's check problem 18 quickly? The user asked for both? Looking back, the image has two problems: 17 and 18.

The instruction says "solve the problem", but there are two. Probably should do both.

Problem 18: Find m∠DGF to nearest whole degree.

Given triangle DEF? Points D, E, F, G.

From diagram: DE = 30, EF = 20, DF = 35? Wait, no.

Looking: D to E is 30, E to F is 20? But there's a brace saying 20 over EG and GF? Wait.

Actually, it says: from E to G to F, and a brace labeled 20, meaning EG + GF = 20? Or is it EF = 20?

Looking carefully: in the diagram, it shows E connected to F, and G is on EF? And there's a brace from E to F labeled 20, so EF = 20.

Also, DG = 35? It says 35 on DG.

And angle at E is 120°.

Points: D, E, F form a triangle? With G on EF.

Given: DE = 30, EF = 20, angle DEF = 120°, and DG = 35. Need angle DGF.

Angle DGF is at G in triangle DGF.

First, find coordinates again.

Place point E at origin (0,0). Since angle at E is 120°, and DE = 30, EF = 20.

Assume ED is along the x-axis? Or better: place E at (0,0), and since angle at E is between DE and FE, set DE along positive x-axis.

So point D: since DE = 30, and if we put D at (30, 0)? But angle at E is 120°, so if DE is along positive x-axis, then EF is at 120° from it.

Standard: angle at E is angle between vectors ED and EF.

If we place E at (0,0), and D at (30, 0), then vector ED is (30,0), so direction 0°.

Then vector EF makes 120° with ED, so EF is at 120° from positive x-axis.

Length EF = 20, so point F is at (20 * cos(120°), 20 * sin(120°))

cos(120°) = -0.5, sin(120°) = √3/2 ≈ 0.8660

So F: (20 * -0.5, 20 * 0.8660) = (-10, 17.32)

Point D: (30, 0)

Now, G is on EF. Since EF is from E(0,0) to F(-10,17.32), and we don't know where G is, but we know DG = 35.

Let G divide EF in some ratio. Parametrize G.

Let G be a point on EF: so G = t * F + (1-t) * E = t*(-10, 17.32) for t in [0,1]

So G = (-10t, 17.32t)

Distance from D(30,0) to G(-10t, 17.32t) is 35.

So:

√[ (30 - (-10t))² + (0 - 17.32t)² ] = 35

Simplify:

√[ (30 + 10t)² + (-17.32t)² ] = 35

Square both sides:

(30 + 10t)² + (17.32t)² = 1225

Compute:

(30 + 10t)² = 900 + 600t + 100t²
(17.32t)² ≈ 300 t² (since 17.32² = 300 approximately? 17.32 * 17.32.

17^2=289, 0.32*34≈10.88, 0.32^2≈0.1024, total ≈ 289+10.88+0.1024=299.9824 ≈ 300)

So:

900 + 600t + 100t² + 300t² = 1225
400t² + 600t + 900 = 1225
400t² + 600t - 325 = 0

Divide by 25: 16t² + 24t - 13 = 0

Solve quadratic: t = [-24 ± √(24² - 4*16*(-13))] / (2*16)

Discriminant: 576 + 832 = 1408

√1408 = √(64*22) = 8√22 ≈ 8*4.6904 ≈ 37.5232

So t = [-24 + 37.5232]/32 ≈ 13.5232/32 ≈ 0.4226 (take positive since t>0)

Or t = [-24 -37.5232]/32 <0 discard.

So t ≈ 0.4226

Thus G is at (-10*0.4226, 17.32*0.4226) ≈ (-4.226, 7.32)

Now, we need angle DGF, which is angle at G in triangle DGF.

Points: D(30,0), G(-4.226,7.32), F(-10,17.32)

Vectors from G:

Vector GD = D - G = (30 - (-4.226), 0 - 7.32) = (34.226, -7.32)
Vector GF = F - G = (-10 - (-4.226), 17.32 - 7.32) = (-5.774, 10)

Angle at G is between vectors GD and GF.

Dot product: GD • GF = (34.226)*(-5.774) + (-7.32)*(10)

Compute:

34.226 * -5.774 ≈ -197.6 (let me calculate: 34*5.774≈196.316, 0.226*5.774≈1.305, total ≈197.621, so -197.621)

-7.32 * 10 = -73.2

Total dot product ≈ -197.621 -73.2 = -270.821

Magnitudes:

|GD| = distance from G to D = 35 (given)
|GF| = distance from G to F: since G is on EF, and EF=20, and t=0.4226, so EG = t * EF? In our parametrization, when t=0, G=E, t=1, G=F, so distance from E to G is t * |EF| = 0.4226 * 20 ≈ 8.452, so GF = EF - EG = 20 - 8.452 = 11.548

Confirm with coordinates: G(-4.226,7.32), F(-10,17.32)

dx = -10 - (-4.226) = -5.774
dy = 17.32 - 7.32 = 10
So |GF| = √[(-5.774)^2 + 10^2] = √[33.34 + 100] = √133.34 ≈ 11.548 — yes.

So |GD| = 35, |GF| ≈ 11.548

Dot product ≈ -270.821

So cos(angle) = dot product / (|GD| |GF|) = -270.821 / (35 * 11.548) ≈ -270.821 / 404.18 ≈ -0.6700

Then angle = arccos(-0.6700) ≈ ?

cos(132°) = -cos(48°) ≈ -0.6694 — very close.

So angle ≈ 132°

To nearest whole degree: 132°

But let me confirm calculation.

35 * 11.548 = 35*11 = 385, 35*0.548=19.18, total 404.18 — ok.

Dot product: GD • GF = (34.226)(-5.774) + (-7.32)(10)

34.226 * 5.774:

30*5.774=173.22, 4.226*5.774≈4*5.774=23.096, 0.226*5.774≈1.305, total ≈23.096+1.305=24.401, so total 173.22+24.401=197.621, so negative -197.621

Plus -73.2 = -270.821 — ok.

-270.821 / 404.18 ≈ -0.6700

arccos(-0.6700): cos(132°) = -cos(48°) = -0.6694, cos(132.1°) = -cos(47.9°) ≈ -0.6704? Let's see.

Actually, cos(48°) = 0.6694, so -0.6694 for 132°.

Our value is -0.6700, which is slightly more negative, so angle slightly larger than 132°.

Difference: 0.6700 - 0.6694 = 0.0006

Derivative: d(cos)/dtheta = -sin, sin(132°)=sin(48°)≈0.7431

So dtheta ≈ -d(cos)/sin ≈ -(-0.0006)/0.7431 ≈ 0.000807 radians ≈ 0.046 degrees

So angle ≈ 132° + 0.046° ≈ 132.05° — still rounds to 132°

So for problem 18, answer is 132°

But the user might expect only one problem? Looking back, the initial request says "solve the problem", but image has two. Since both are labeled, probably do both.

In the response, I'll provide both answers.

Final Answer for 17: 39°
Final Answer for 18: 132°

But the format says "Final Answer" section, so perhaps list both.

Since the problems are separate, I'll state them clearly.

──────────────────────────────────────

Final Answer:
For problem 17: 39°
For problem 18: 132°