Law of Sines and Cosines Worksheet with Key (pdf). - Free Printable

Let’s solve problem 17 first.

We are given triangle ABC with point D on side AB. We know:

- AC = 35
- CD = 25
- DB = 12
- Angle CDB = 118° (this is the angle at D in triangle CDB)

We need to find angle A (angle CAB), to the nearest whole degree.

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Step 1: Look at triangle CDB.

In triangle CDB, we know two sides and the included angle? Wait — actually, we know:

- Side CD = 25
- Side DB = 12
- Angle at D = 118° → that’s angle CDB

So yes, 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.

Let’s compute:

CB² = 25² + 12² - 2·25·12·cos(118°)

First, calculate squares:

25² = 625
12² = 144
Sum = 625 + 144 = 769

Now, cos(118°). Let’s get that value.

cos(118°) ≈ cos(180° - 62°) = -cos(62°) ≈ -0.4695

So:

-2·25·12·(-0.4695) = +2·25·12·0.4695

Calculate step by step:

2 × 25 = 50
50 × 12 = 600
600 × 0.4695 ≈ 281.7

So CB² ≈ 769 + 281.7 = 1050.7

Then CB ≈ √1050.7 ≈ 32.41

Okay, so CB ≈ 32.41

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Step 2: Now look at triangle ACD or triangle ABC?

Actually, we want angle A, which is in triangle ACD or triangle ABC.

Note: Points A, D, B are colinear, with D between A and B.

We don’t know AD yet. But we do know AC = 35, CD = 25, and if we can find angle ADC, then we could use Law of Sines or Cosines in triangle ACD.

Wait — angle CDB is 118°, and since A-D-B is a straight line, angle ADC = 180° - 118° = 62°

Yes! Because angles on a straight line add to 180°.

So in triangle ACD, we now know:

- AC = 35
- CD = 25
- angle ADC = 62°

And we want angle at A (angle CAD).

Perfect — we can use Law of Sines in triangle ACD.

Law of Sines:

sin(angle A) / CD = sin(angle ADC) / AC

That is:

sin(A) / 25 = sin(62°) / 35

Solve for sin(A):

sin(A) = 25 · sin(62°) / 35

Compute sin(62°) ≈ 0.8829

So:

sin(A) = 25 × 0.8829 / 35 ≈ 22.0725 / 35 ≈ 0.6306

Now take arcsin:

A ≈ arcsin(0.6306) ≈ ?

Using calculator: arcsin(0.6306) ≈ 39.1°

To nearest whole degree: 39°

Wait — let me double-check calculations.

Check:

sin(62°) = 0.88294759285...

25 * 0.88294759285 = 22.07368982125

Divide by 35: 22.07368982125 / 35 = 0.6306768520357

arcsin(0.6306768520357) = ?

Use calculator: sin⁻¹(0.6307) ≈ 39.1 degrees → rounds to 39°

Is this correct? Let’s verify using Law of Cosines in triangle ACD to check angle A.

In triangle ACD:

Sides: AC=35, CD=25, angle at D=62°

We can find side AD using Law of Cosines:

AC² = AD² + CD² - 2·AD·CD·cos(angle ADC)

Wait — no, better to use Law of Cosines to find angle at A directly.

Law of Cosines for angle A:

cos(A) = (AC² + AD² - CD²) / (2·AC·AD)

But we don’t know AD.

Alternatively, use Law of Sines as before — it should be fine.

Another way: Use Law of Cosines to find AD first.

In triangle ACD:

CD² = AC² + AD² - 2·AC·AD·cos(angle A)

No, that’s circular.

Better: Use Law of Cosines to find side AD.

From triangle ACD:

We have sides AC=35, CD=25, angle at D=62°

So, by Law of Cosines:

AC² = AD² + CD² - 2·AD·CD·cos(angle ADC)

Wait — angle at D is between sides AD and CD, so yes:

AC² = AD² + CD² - 2·AD·CD·cos(62°)

Plug in:

35² = AD² + 25² - 2·AD·25·cos(62°)

1225 = AD² + 625 - 50·AD·cos(62°)

cos(62°) ≈ 0.4695

So:

1225 = AD² + 625 - 50·AD·0.4695

1225 = AD² + 625 - 23.475·AD

Bring all to one side:

AD² - 23.475·AD + 625 - 1225 = 0

AD² - 23.475·AD - 600 = 0

Now solve quadratic:

AD = [23.475 ± √(23.475² + 4·600)] / 2

Compute discriminant:

23.475² ≈ 551.075625

4·600 = 2400

Sum ≈ 551.075625 + 2400 = 2951.075625

√2951.075625 ≈ 54.325

So AD = [23.475 + 54.325]/2 ≈ 77.8/2 ≈ 38.9 (take positive root since length)

Or [23.475 - 54.325]/2 negative → discard

So AD ≈ 38.9

Now, in triangle ACD, use Law of Cosines to find angle A:

cos(A) = (AC² + AD² - CD²) / (2·AC·AD)

= (35² + 38.9² - 25²) / (2·35·38.9)

Compute numerator:

35² = 1225
38.9² ≈ 1513.21
25² = 625

So: 1225 + 1513.21 - 625 = (1225 - 625) + 1513.21 = 600 + 1513.21 = 2113.21

Denominator: 2·35·38.9 = 70·38.9 = 2723

So cos(A) ≈ 2113.21 / 2723 ≈ 0.7761

Then angle A ≈ arccos(0.7761) ≈ ?

arccos(0.7761) ≈ 39.1° → again 39°

Same answer!

So angle A is approximately 39 degrees.

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Final Answer:
39