​Solving a non–right triangle via Law of Sines and Cosines

Preliminary words on the Law of Sines and Cosines:

1. First, sketch and label the sides and angles of the triangle.

2. Identify the triangle rule that applies: ASA, AAS, ASS, SAS or SSS

3. For ASA, AAS and ASS, use Law of Sines:

sin A/ a = sin B / b = sin C / c

Sometimes, this is expressed as its reciprocal:

a / sin A = b / sin B = c / sin C

Dont MEMORIZE the formulas –– only remember the pattern: it's basically the RATIOS of the sin of the angles of any angle of a triangle and the side across from the angle are PROPORTIONAL!

4. In case of Ambiguous Case [SSA], upon finding the sin of the angle, remember that the result is the reference angle: find the acute [1st quadrant angle ~ reference angle] and obtuse angle [2nd quadrant angle ~ 180 – Reference Angle] that satisfies the equation. Find the 3rd pairs of angles using: Triangle Sum = 180 degrees.
Then, find the 3rd side(s).

5. For SAS, use Law of Cosines: In any triangle, with 2 sides and an included angle, first, find the missing side using:

a² =  b² + c² – 2bc cos A

Remember to take square–roots!

Dont MEMORIZE the formulas –– only remember the pattern.

6. For SSS, use Law of Cosines: In any triangle, with 3 known sides, find the missing angles using:

cos A =  (b² + c² – a²)  / 2bc

Use ( ) in both numerator and denominator!

Dont MEMORIZE the formulas –– only remember the pattern:

I Law of Sines
To use the Law of Sines you need to know either two angles and one side of the triangle (AAS or ASA) or two sides and an angle opposite one of them (SSA).  Notice that for the first two cases we use the same parts that we used to prove congruence of triangles in geometry but in the last case we could not prove congruent triangles given these parts. This is because the remaining pieces could have been different sizes.  This is called the ambiguous case.



According to the law,

where a, b, and c are the lengths of the sides of a triangle, and A, B, and C are the opposite angles (see the figure to the right), and D is the diameter of the triangle's circumcircle. When the last part of the equation is not used, sometimes the law is stated using the reciprocal:

Proof of the Law of Sines


Considering the left right–triangle, if h is the height, since sin A = h / b, the height, h = b sin A.
Similarly, observing the right right–triangle, sin B = h / a, the height, h = a sin B.

Since both are expressions for the height, h: b sin A = a sin B.

Cross–multiplying, sin A / a = sin B / b

If we dropped the perpendicular from A, we could similarly prove: 

b sin C = c sin B yielding sin B / b = sin C / c

so that combining:

Example 1: Given two angles and a non–included side (AAS).

Given ∆ABC with A = 30°, B = 20° and a = 45 m.  Find the remaining angle and sides.

       The third angle of the triangle is

              C = 180° – A – B = 180° – 30° – 20 ° = 130°

        By the Law of Sines,

       By the Properties of Proportions

Example 2: Given two angles and an included side (ASA).

Given A = 42°, B = 75° and c = 22 cm.  Find the remaining angle and sides.

       The third angle of the triangle is:

              C = 180° – A – B = 180° – 42° – 75° = 63°

       By the Law of Sines,

       By the Properties of Proportions

               and

The Ambiguous Case

If two sides and an angle opposite one of them is given, three possibilities can occur.

            (1) No such triangle exists.

            (2) Two different triangles exist.

            (3) Exactly one triangle exists.

Consider a triangle in which you are given a, b and A.  (The altitude h from vertex B to side

​ AC, by the definition of sines is equal to b sin A.)

            (1) No such triangle exists if A is acute and a < h or A is obtuse and a ≤ b.

            (2) Two different triangles exist if A is acute and h < a < b.

            (3) In every other case, exactly one triangle exists.

Relax! Which of the 3 cases a triangle in the Ambiguous Case shall be...shall easily work itself out, naturally! There's nothing to "memorize".

Example 1: No Solution Exists

Given a = 15, b = 25 and A = 80°.  Find the other angles and side. 

              h = b sin A = 25 sin 80° ≈ 24.6

       Notice that a < h. So it appears that there is no solution.  Verify this using the Law of Sines.

This contrasts the fact that the –1 ≤ sin B ≤ 1.  Therefore, no triangle exists because when you attempt to find sin^–1(1.641) youll get an ERROR!

Bottomline: it worked out that there are no triangles that are possible! There are no "rules" to memorize!

Example 2: Two Solutions Exist

Given a = 6. b = 7 and A = 30°.  Find the other angles and side. 

              h = b sin A = 7 sin 30° = 3.5

              h < a < b therefore, there are two triangles possible.

       By the Law of Sines,

There are two angles between 0° and 180° whose sine is approximately 0.5833, 35.69° and 144.31°: because if sin B = 0.5833, then the ratio being positive, angle B can be (acute) in the 1st quadrant, B = sin^–1(0.64958...) = 35.69° or angle B can be (obtuse) in the 2nd quadrant, 180 – 35.69° = 144.31°.

                  If B ≈ 35.69°                                               If B ≈ 144.31°

Bottomline: it worked out that there are 2 triangles that are possible! There are no "rules" to memorize!

Therefore, there are 2 possibilities for angle C:

                  C ≈180° – 30° – 35.69° ≈ 114.31°             C ≈ 180° – 30° – 144.31° ≈ 5.69°

Example 3: One Solution Exists

Given a = 22, b =12 and A = 40°.  Find the other angles and side.

                 a > b

       By the Law of Sines,

There are two angles between 0° and 180° whose sine is approximately 0.3506, 20.52° and 159.48°:

because if sin B = 0.3506, then the ratio being positive, angle B can be (acute) in the 1st quadrant, B = sin^–1(0.3506...) = 20.52° or angle B can be (obtuse) in the 2nd quadrant, 180 – 20.52° = 159.48°.

                  If B ≈ 20.52°                                               If B ≈ 159.48°

Therefore, there are 2 possibilities for angle C:

             C ≈ 180° – 40° – 20.52° ≈ 119.48°.             C ≈ 180° – 40° –159.48°  ≈ –19.48° which is impossible!

Bottomline: it worked out that there's only the 1st triangle that is possible! There are no "rules" to memorize!

       By the Law of Sines,

II Law of Cosines
If we are given two sides and an included angle of a triangle (SAS) or if we are given 3 sides of a triangle (SSS), we cannot use the Law of Sines because we cannot set up any proportions where enough information is known.  In these two cases we must use the Law of Cosines.

The Law of Cosines states:

       c² = a² + b² – 2abcos C.

This resembles the Pythagorean Theorem except for the third term and if C is a right angle the third term equals 0 because the cosine of 90° is 0 and we get the Pythagorean Theorem.  So, the Pythagorean Theorem is a special case of the Law of Cosines.

The Law of Cosines can also be stated as

       b² = a² + c² – 2accos B or

       a² = b² + c² – 2bccos A.

Proof of the Law of Cosines
The Law of Cosines states that given any triangle with side lengths a, b and c and opposing angles A, B and C: c² = a² + b² − 2ab cos C​

​For this, sketch triangle ABC with angle B on top, angle A on the bottom right and angle C on the bottom left. ​

First we'll divide the triangle into two right triangles by dropping the perpendicular​ h from angle B (top vertex) so that h ​​is perpendicular to side b.

​
[See outline here: http://mathproofs.blogspot.com/2006/06/law–of–cosines.html but the explanations below are more detailed!]​

If h is the height, since sin C = h / a, the height, h = a sin C.

Similarly, if x is  the base of the left right–triangle, cos C = x / a so that: x = a cos C so that the base of the right right–triangle is, b – a

​cos C since b ~ entire base.

​Now using Pythagoras' Theorem on the bottom right triangle with c as the hypotenuse:

c² = (a sin C)² + (b − a cos C​)²

​=> ​c² = a² sin²C + b² − 2ab cos C + a² cos²C

​=> ​c² = a² (sin²C + cos²C) + b² − 2ab cos C

​=> ​c² = a² (1) + b² − 2ab cos C since sin²C + cos²C = 1
=> c² = a² + b² − 2ab cos C

Here's a similar approach:

Let ABC be a triangle with sides a, b, c.  We will show

​​

​​

c² = a² + b² − 2ab cos C.

(The trigonometric functions are defined in terms of a right–angled triangle. Therefore it is only with the aid of right–angled triangles that we can prove anything)

Draw BD perpendicular to CA, separating triangle ABC into the two right triangles BDC, BDA.  BD is the height h of triangle ABC.

Call CD x.  Then DA is the whole b minus the segment x:  b − x.

Also, since

then

x  =  a cos C .  .  .  .  .  .  . (1)

Now, in the right triangle BDC, according to the Pythagorean theorem:

h² + x² = a²,

so that

h² = a² − x².  .  .  .  .  .  (2)

In the right triangle BDA,

c² = h² + (b − x)²

c² = h² + b² − 2bx + x².

For h², let us substitute line (2):

c² = a² − x² + b² − 2bx + x²

c² = a² + b² − 2bx.

Finally, for x, let us substitute line (1):

c² = a² + b² − 2b· a cos C.

That is,

c² = a² + b² − 2ab cos C.

This is what we wanted to prove.

In the same way, we could prove that

a² = b² + c² − 2bc cos A

and

b² = a² + c² − 2ac cos B.

This is the Law of Cosines.

Alternately, consider:

Using reasoning similar to that before:

(This is still true if α or β is obtuse, in which case the perpendicular falls outside the triangle.) Multiply through by c to get

By considering the other perpendiculars obtain

Adding the latter two equations gives

Subtracting the first equation from the last one we have

which simplifies to

Find here proofs for both, the Law of Sines and Cosines.

Example 1: Two Sides and the Included Angle–SAS

Given a = 11, b = 5 and C = 20°.  Find the remaining side and angles.

To find the remaining angles, it is easiest to now use the Law of Sines.

Example 2: Three Sides–SSS

Given a = 8, b = 19 and c = 14.  Find the measures of the angles.

It is best to find the angle opposite the longest side first.  In this case, that is side b.

       Because cos B is negative, we know that B is an obtuse angle.

              B ≈ 116.80°

Since B is an obtuse angle and a triangle has at most one obtuse angle, we know that angle A and angle C are both acute.

       To find the other two angles, it is simplest to use the Law of Sines. 

Solved Examples
1. What is the length of side a?

Solution. Use the Law of Sines a/sin A = b/sin B = c/sin C

We know angle A = 54°, side b = 3 and angle B = 27°

Substitute these values into a/sin A = b/sin B

Therefore a/sin 54° = 3/sin 27°

Therefore a = (3 × sin 54°)/sin 27° = 5.346 correct to 3 decimal places

2. What is the length of side b?


Solution. First use angles of a triangle to find the size of angle C:
C = 180° – 69° – 32° = 79°

Now use the Law of Sines: a/sin A = b/sin B = c/sin C

We know angle B = 32°, side c = 5.7 and angle C = 79°

Substitute these values into b/sin B = c/sin C

Therefore b/sin(32°) = 5.7/sin(79°)

Therefore b = sin(32°) × 5.7/sin(79°) = 3.08 correct to 2 decimal places

3. What is the length of side c?


Solution. First use angles of a triangle to find the size of angle C
C = 180° – 50° – 22° = 108°

Now use the Law of Sines: a/sin A = b/sin B = c/sin C

We know angle B = 22°, side b = 3.2 and angle C = 108°

Substitute these values into b/sin B = c/sin C

Therefore 3.2/sin(22°) = c/sin(108°)

Therefore c = 3.2 × sin(108°)/sin(22°) = 8.12 correct to 2 decimal places.

4. What is the size of Angle C?


Solution. Use the Law of Sines  sin(A)/a = sin(B)/b = sin(C)/c

We know angle B = 77°, side b = 9 and side c = 6

Substitute these values into sin(B)/b = sin(C)/c

Therefore sin(77°)/9 = sin(C)/6

Therefore sin(C) = ( sin(77°)/9 ) × 6 = 0.64958...

Therefore C = sin^–1(0.64958...) = 40.5°.

But being the Ambiguous Case, you must consider angle C being obtuse: C = 180  – 40.5° =  139.5° which makes the 3rd angle A, negative for this possibility: A = 180  – 139.5° – 70° < 0!
 
5. What is the size of Angle X?


Solution. Use the Law of Sines  sin(X)/x = sin(Y)/y = sin(Z)/z

We know angle Y = 108°, side y = 11.7 and side z = 5.6

Substitute these values into sin(Y)/y = sin(Z)/z

Therefore sin(108°)/11.7 = sin(Z)/5.6

Therefore sin(Z) = (5.6 × sin(108°))/11.7 = 0.45520...

Therefore Z = sin^–1(0.45520...) = 27.1°.

But being the Ambiguous Case, you must consider angle Z being obtuse: Z = 180  – 27.1° =  152.9° which makes the 3rd angle, X, negative for this possibility: X = 180  – 152.9° – 108° < 0!

Now use angles of a triangle to find X:

X = 180° – 108° – 27.1° = 44.9°.

6. What is the size of Angle P?

Solution. Use the Law of Sines  sinP/p = sinQ/q = sinR/r

We know angle R = 36°, side r = 7.4 and side p = 12.5

Substitute these values into sinP/p = sinR/r

Therefore sinP/12.5 = sin36°/7.4

Therefore sinP = (12.5 × sin36°)/7.4 = 0.99288...

Therefore P = sin^–1(0.99288...) = 83.2° or 96.8°

Remember there can be two answers in some cases, and this is one of them, where the two answers differ by 180° because the sine of an angle is positive in, both, the 1st and the 2nd quadrants!

In this case, the two answers are 83.2° and (180 – 83.2)° = 96.8°
So you should give both answers. Answers A and B are not wrong, but they each only give one of the two possible answers.

7. Farmer Jones has a triangular field ABC as shown in the following diagram. He wants to fence the field with the fencing is sold by the meter. How many meters of fencing does he need to purchase?

Solution. We need to find the lengths of the two remaining sides, so that we can find the Perimeter of the field:

Use the Law of Sines a/sin A = b/sin B = c/sin C.

First find the length of AB = c.
We know angle C = 114°, side b = 300 and angle B = 27°.

Substitute these values into c/sin C = b/sin B.

Then c/sin 114° = 300/sin 27°.
Next c = (300 × sin 114°)/sin 27° = 603.7 correct to 1 decimal place.

Angle A = 180° – (114° + 27°) = 39°.
Next find the length of BC = a.
We know angle A = 39°, side b = 300 and angle B = 27°.

Substitute these values into a/sin A = b/sin B.

So a/sin 39° = 300/sin 27°.

Then a = (300 × sin 39°)/sin 27° = 415.9 correct to 1 decimal place.

So, the perimeter of the field = 300 m + 603.7 m + 415.9 m = 1,319.6 m,

and Farmer Jones needs to purchase 1,320 meters of fencing.

8. Mrs Jones goes on a round trip from Town A to Town B to Town C and back to Town A, as shown in the following diagram.

All roads are straight. To the nearest mile. How long is the round trip?

Solution. First use the Law of Sines  sin(A)/a = sin(C)/c to find angle A.

We know angle C = 63°, side c = 46 and side a = 34.

Substitute these values into sin(A)/a = sin(C)/c.

Then sin(A)/34 = Sin(63°)/46

So sin(A) = ( sin(63°)/46 ) × 34 = 0.6585...

Then A = sin^–1(0.6585...) = 41.2°.

Next find angle B using angles of a triangle add to 180°
⇒ B = 180° – (63° + 41.2°) = 75.8°

Next use the Law of Sines again to find side b, this time using b/sinB = c/sinC.

We know angle C = 63°, side c = 46 and angle B = 75.8°.

Substitute these values into b/sinB = c/sinC.

Then b/sin(75.8°) = 46/sin(63°)

So b = 46/sin(63°) × sin(75.8°) = 50.04.

The round trip, therefore = 46 miles + 34 miles + 50.04 miles = 130 miles to the nearest mile.

9. Which is the area of the triangle?

Give your answer in square units correct to 1 decimal place.

Solution. Notice that we are given two sides, but not the INCLUDED angle.
Then we need to calculate some other angles and/or sides first, using the Law of Sines.

First use sin(A)/a = sin(C)/c to find angle A.

We know angle C = 30°, side c = 7 and side a = 10.

Substitute these values into sin(A)/a = sin(C)/c.

So sin(A)/10 = Sin(30°)/7.

Then sin(A) = ( sin(30°)/7) × 10 = 0.7142...

Therefore A = sin^–1(0.7142...) = 45.6°
However, A is clearly obtuse ⇒ A = 180 – 45.6° = 134.4°

Next find angle B using angles of a triangle add to 180°
⇒ B = 180° – (30° + 134.4°) = 15.6°

We now know two sides and the included angle: a = 10 and c = 7, and B = 15.6°

Then we use the formula Area = ½ ac sin B.

Area = ½ ac sin B
= ½ × 10 × 7 × sin15.6°
= 35 × sin15.6°
= 35 × 0.2686...
= 9.4 correct to 1 decimal place

10. An observer stands at the summit M of a mountain. The point N is at the base of the mountain vertically below M. The observer sees a lake L at an angle of depression of 15° and a knoll K at an angle of depression of 32°.
The points N, K and L all lie in a straight line and the distance between L and K is 4 miles. Which is the height of the mountain?



Solution. In triangle MLK, we know m = 4, L = 15°.
Also M = 32° – 15° = 17°.
So triangle MLK is an AAS triangle:


Now find side l using the Law of Sines, l/sin L = m/sin M:
Then l/sin 15° = 4/sin 17°
So l = (4 × sin15°)/sin 17° = 3.54 correct to 2 decimal places.


Now use trigonometry again in right triangle KMN to find h, the height of the mountain:

Sin 32° = h/3.54 ⇒ h = 3.54 × sin 32° = 1.876...

The height of the mountain is 1.88 miles correct to 2 decimal places.
(That's about 9,900 feet.)

11.



Find the length of BC.

Solution. Step 1 Find angle C using "angles of a triangle add to 180 degrees":

C = 180° – (107° + 28°)
= 180° – 135°
= 45°

Step 2 Use the Law of Sines, a/sin A = c/sin C, to find side a:
Then a/sin 28° = 12.6/sin45°.
So a = (12.6 × sin 28°)/sin 45° = 8.366 correct to 3 decimal places.

12.


Find the length of side a.

Solution. Find side a using the Law of Cosines
a² = b² + c² – 2bc cos A
= 4.5² + 7.2² – 2 × 4.5 × 7.2 × cos129°
= 20.25 + 51.84 – 64.8 × cos129°
= 72.09 – 64.8 × (–0.6293...)
= 72.09 – (–40.7799...)
= 72.09 + 40.7799...
= 112.8699...
So b = √112.8699... = 10.62 correct to 2 decimal places.

13. In triangle ABC,  DA = DB = DC (see the figure).



Find the value of angle ACB.


Solution.

Triangle ACD is isosceles. Let angle ACD = angle CAD = x°
and, because angles of a triangle add to 180° ⇒ angle ADC = (180 – 2x)°

Triangle CDB is isosceles. Let angle DCB = angle CDB = y°
and, because angles of a triangle add to 180° ⇒ angle BDC = (180 – 2y)°



But adjacent angles on a straight line also add to 180°
⇒ angle ADC + angle BDC = 180°
⇒ (180 – 2x) + (180 – 2y) = 180
⇒ 360 – 2(x + y) = 180
⇒ 2(x + y) = 180
⇒ x + y = 90
⇒ Angle ACB is right.

Alternatively, use the fact that the angle in a semicircle is 90°:

14. The diagram show two swimmers R and S on the sea surface looking at a point P on the seabed. The distance from R to P is 120 feet and the angle of depression from R to P is 32°. The distance from S to P is 95 feet. Calculate the angle of depression, θ, from S to P.

Solution.

In triangle PRS, we know r = 95, s = 120 and R = 32°
Therefore PRS is an SSA (Side,Side,Angle) triangle and we can find S by using the Law of Sines:

sin(S)/s = sin(R)/r:
⇒ sin(S)/120 = sin32°/95
⇒ sin(S) = (120 × sin(32°))/95 = (120 × 0.5299...)/95 = 0.6693...
⇒ S = sin^–1(0.6693...) = 42.0° (by my calculator)
BUT the angle S is actually the obtuse angle: ⇒ S = 180 ° – 42.0° = 138.0°
And the angle of depression from S to P is the acute angle θ, which is 42.0°

15. A cylindrical container of height 32 cm and base diameter 18 cm is filled with sand. The container is emptied on the ground and a conical heap of sand is formed. If the height of the conical heap is 24 cm, find the radius of the heap.

Solution. Volume of the cylinder

where r is the radius and h is the height.

Volume of a cone is


The cylindrical container has height 32 cm and diameter 18 cm. Since diameter is 18 cm, the radius is 9 cm.
Therefore, volume of the cylinder is


The volume of the cone would be the same as the cylindrical container as the same quantity of sand occupied the cylindrical container and is then in the shape of a cone. The height can be calculated with the help of the volume. The height of the conical heap is given as 24 cm.



Cancelling pi on both the sides and rearranging the terms,

Taking square root on both sides,


Therefore, the radius of the cone is 18 cm.

16. A cylindrical tunnel 5.5m in radius is driven through a mountain for a distance of 7km. The excavated earth is spread to a depth of 50cm on the surface of a road 11m wide. Calculate the number of kilometers of the road that can be covered with all of the excavated earth. Use (22/7) as an approximation for π.

Solution. The volume of the cylindrical tunnel is: π × 5.5² × 7000 = 665,500 m³ (Taking pi as 22/7)

That volume spread out over 0.5m × 11m is:

665,500/(0.5m × 11m) = 121000 m, or 121 km

17. A cylindrical jar of 15 cm diameter contains milk to a height of 6 cm. The milk is poured into conical bottles of radius 1.25 cm and height 6 cm.  How many of those conical bottles can be filled completely with milk from the jar?

Solution. Let V₁ be the volume of milk in the cylindrical jar.

where r = 7.5, h = 6.


Let the volume of the conical bottle be V₂.

r = 1.25 cm and h = 6cm.


Dividing the two:



Hence, 108 conical bottles can be filled completely with milk.

18. Solve triangle ABC:

Solution.
Step 1 A is the largest angle, so find A first using the Law of Cosines: cosA = (b² + c² – a²)/2bc
Then cosA = (2.9² + 3.7² – 6.4²)/(2 × 2.9 × 3.7)
= (8.41 + 13.69 – 40.96)/21.46
= –18.86/21.46
= –0.8788...
So A = 151.5°.

Step 2 Use the Law of Sines, sinB/b = sinA/a, to find angle B:

Then sinB/2.9 = sin151.5°/6.4.
Next sinB = (2.9 × sin151.5°)/6.4 = 0.2161...
So B = 12.5°.

Step 3 Find angle C using 'angles of a triangle add to 180°':

C = 180° – (151.5° + 12.5°)
= 180° – 164.0°
= 16.0°

Finally A = 151.5°,  B = 12.5° and C = 16.0°.

19. The diameters of the bases of two cones are equal. If their slant heights are in the ratio 5:4, find the ratio of their curved surface areas.

Solution.
The curved Surface Area of a cone is given by the formula:

where r is the radius of the cone and s is the slant height.
In the problem given, it is stated that the diameters of the bases of the two cones are equal. Hence their radii are equal too.
The ratio of the slant heights is 5:4
Therefore, the ratio of the curved surface areas would be:


It is also given in the problem that the ratio of the slant heights is 5:4
Therefore


Hence, the curved surface areas of the two cones are in the ratio 5:4

20. More solved examples on the Law of Sines and Cosines: here and here. Try them out yourselves!