# The Law of Sines – Proof & How it works?

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Law of Sines is helpful in solving any triangle with certain requirements like the side or angle must be given in order to proceed with this law. This law considers ASA, AAS, or SSA. which is one case because knowing any two angles & one side means knowing all the three angles & one side.

Note: Law of Cosine doesn’t work where at least two sides are needed.

This is the formula for The Law of Sines

where,

side “a” faces angle “α”.
Similarly, side “b” faces angle “β”.
side “c” faces angle “γ”.

The law states that if we divide side “a” by the sine of angle “α” is equal to side “b” divided by the sine of angle of “B” & side “c” divided by the sine of angle of “γ”.

When any one part of the side & angle is given say (side a & angle A)

Using the Law of Sines as stated above,
a/ Sine α —> (1)

This result obtained from (1) is equal to result obtained from (2) & (3).

b/ Sine β —> (2)

c / Sine γ —> (3)

which means you can try & use any one part of the corresponding sides & angles will give you exactly the same answer.

Let’s try out a problem based on this law,

Q1: Solve the triangle ABC, given a = 30, b = 70, β = 85°. Find α.

Solution: Since it is an SSA related question.

Using the Law of Sines,

sin α / a = sin β / b
sin α / 30 = sin 85° / 70
sin α = 30▪︎sin 85°/70
sin α = 0.4269
α = sin-1▪︎(04269)
α = 25°

Q2. Solve the triangle ABC in which α = 38°, β = 121° and a = 20.

Solution: Since it is an AAS related case,

Using the Law of Sines,

b / sin β = a / sin α
b / sin 121° = 20 / sin 38°
b = a▪︎sin β / sin α
b = 28 approximately.

Using the Law of Sines again but this time using α, γ to get c.

c / sin γ = a / sin α
c / sin 21° = 20 / sin 38°
c = 20▪︎sin 21° / sin 38°
c = 11.6 ~ 12 approximately.

Here is a word problem for you to enhance your comprehension.

Q3. Solve the triangle ABC where β = 39°30′ , γ = 34°10′, c = 115.

Solution: Since it is an AAS related case,

Given: a= 240, β = 39.5° and γ = 34.16°

α = 180° – (β + γ)
= 180° – 73.66°
= 106.34°

Using the Law of Sines,

sin α / a = sin β / b
b = sin β * a / sin α
b = 39.5°
b = (0.63)*(240) / (0.95)
b = 151.2 / .95
b = 159.15

Now, to get side “c”, just use Pythagoras theorem and you are good to go.

c² = a² + b²
c² = 57600 + 25328.72
√c² = 82928.72
c = 287.97

Q4. From a point A, the angle of elevation of the top C of a tower is 28°. From a second point B, which is 2200 ft closer to the base of the tower, the angle of elevation of the top is 66°. What is the height “h” of the tower?

Solution: First we need to work out on the triangle because we may end up getting confused & wrong answer.

Given: For triangle ABC,
AB = 2200 ft
α = 28°
β = 180° – 66° = 114°
γ = 38 °

Applying the Law of Sines to triangle ABC,

sin γ / c = sin α / a
sin 38° / 2200 = sin 28° / a
a = sin 28° X 2200 / sin 38°
a = 1678 ft approximately.

Since we are left with finding out “h” & “x” side with only one angle given in the triangle BDC. Here we use simply trigonometric formula.

sin 66° = h / 1678
h = sin 66°▪︎1678
h = 1533 ft approximately.

I hope you understood the Law of Sines, now you can easily solve any problem related to this law. You can also surprise your friends too by knowing the proof of this law as well.

Theorem: In any triangle ABC with usual labeling sin α / a = sin β / b = sin γ / c.

Proof: Consider any oblique triangle as shown in figure (i) & (ii).

Let h = height of the triangle with base CA. Then in in figure (i)
sin α = h / c
h = a▪︎sin γ

Thus, c▪︎sin α = a▪︎sin γ
sin α / a = sin γ / c –> (1)

In figure (ii),

h = a▪︎sin (180°-C) = a▪︎sin γ
and h = c▪︎sin α

Hence, c▪︎sin α = a▪︎sin γ

Similarly, if we draw perpendicular from the other two vertices on opposite sides of triangle ABC, we get:

sin α / a = sin β / b —> (2)

and sin β / b = sin γ / C —> (3)

Combining equations (1), (2), & (3), we have:

sin α / a = sin β / b = sin γ / c
or equivalently a / sin A = b / sin β = c / sin γ

These equations finally give the Law of Sines.