We will discuss here about the law of sines or the sine rule which is required for solving the problems on triangle.

In any triangle the sides of a triangle are proportional to the sines of the angles opposite to them.

That is in any triangle ABC,

** ****\(\frac{a}{sin A}\) = \(\frac{b}{sin B}\) = \(\frac{c}{sin C}\)**

**Proof:**

Let ABC be a triangle.

Now will derive the three different cases:

**Case I:** Acute angled triangle (three angles are acute): The triangle ABC is acute-angled.

Now, draw AD from A which is perpendicular to BC. Clearly, D lies on BC

Now from the triangle ABD, we have,

sin B = AD/AB

⇒ sin B = AD/c, [Since, AB = c]

⇒ AD= c sin B ……………………………………. (1)

Again from the triangle ACD we have,

sin C = AD/AC

⇒ sin C = AD/b, [Since, AC = b]

⇒ AD = b sin C ...………………………………….. (2)

Now, from (1) and (2) we get,

c sin B = b sin C

⇒ b/sin B = c/sin c………………………………….(3)

Similarly, if we draw a perpendicular to AC from B, we will get

a/sin A = c/sin c………………………………….(4)

Therefore, from (3) and (4) we get,

\(\frac{a}{sin A}\) = \(\frac{b}{sin B}\) = \(\frac{c}{sin C}\)

**Case II:** Obtuse angled triangle (one angle is obtuse): The triangle ABC is obtuse angled.

Now, draw AD from A which is perpendicular to produced BC. Clearly, D lies on produced BC.

Now from the triangle ABD, we have,

sin ∠ABD = AD/AB

⇒ sin (180 - B) = AD/c, [Since ∠ABD = 180 - B and AB = c]

⇒ sin B = AD/c, [Since sin (180 - θ) = sin θ]

⇒ AD = c sin B ……………………………………. (5)

Again, from the triangle ACD, we have,

sin C = AD/AC

⇒ sin C = AD/b, [Since, AC = b]

⇒ AD = b sin C ……………………………………. (6)

Now, from (5) and (6) we get,

c sin B = b sin C

b/sin B = c/sin C ……………………………………. (7)

Similarly, if we draw a perpendicular to AC from B, we will get

a/sin A = b/sin B ……………………………………. (8)

Therefore, from (7) and (8) we get,

\(\frac{a}{sin A}\) = \(\frac{b}{sin B}\) = \(\frac{c}{sin C}\)

**Case III:** Right angled triangle (one angle is right angle): The triangle ABC is right angled. The angle C is a right angle.

Now from triangle ABC, we have,

sin C = sin π/2

⇒ sin C = 1, [Since, sin π/2 = 1], ……………………………………. (9)

sin A = BC/AB

⇒ sin A = a/c, [Since, BC = a and AB = c]

⇒ c = a/sin A ……………………………………. (10)

and sin B = AC/AB

⇒ sin B = b/c, [Since, AC = b and AB = c]

⇒ c = b/sin B ……………………………………. (11)

Now from (10) and (11) we get,

a/sin A = b/sin B = c

⇒ a/sin A = b/sin B = c/1

Now from (9) we get,

⇒ \(\frac{a}{sin A}\) = \(\frac{b}{sin B}\) = \(\frac{c}{sin C}\)

Therefore, from all three cases, we get,

\(\frac{a}{sin A}\) = \(\frac{b}{sin B}\) = \(\frac{c}{sin C}\). *Proved.*

**Note: **

**1.** The sine rule or the law of sines can be expressed as

\(\frac{sin A}{a}\) = \(\frac{sin B}{b}\) = \(\frac{sin C}{c}\)

**2.** The sine rule or the law of sines is a very useful rule to
express sides of a triangle in terms of the sines of angles and vice-versa in
the following manner.

We have \(\frac{a}{sin A}\) = \(\frac{b}{sin B}\) = \(\frac{c}{sin C}\) = k\(_{1}\) (say)

⇒ a = k\(_{1}\) sin A, b = k\(_{1}\) sin B and c = k\(_{1}\) sin C

Similarly, sin A/a = sin B/b = sin C/c = k\(_{2}\) (say)

⇒ sin A = k\(_{2}\) a, sin B = k\(_{2}\) b and sin C = k\(_{2}\) c

Solved problem using the law of sines:

The triangle ABC is isosceles; if ∠A = 108°, find the value of a : b.

**Solution:**

Since the triangle ABC is isosceles and A = 108°, A + B + C = 180°, hence it is evident that B = C.

Now, B + C = 180° - A = 180° - 108°

⇒ 2B = 72° [Since, C = B]

⇒ B = 36°

Again, we have, \(\frac{a}{sin A}\) = \(\frac{b}{sin B}\)

Therefore, \(\frac{a}{b}\) = \(\frac{sin A}{sin B}\) = \(\frac{sin 108°}{sin 36°}\) = \(\frac{cos 18°}{sin 36°}\)

Now, cos 18° = \(\sqrt{1 - sin^{2} 18°}\)

= \(\sqrt{1 - (\frac{\sqrt{5} - 1}{4})^{2}}\)

= ¼\(\sqrt{10 + 2\sqrt{5}}\)

and sin 36° = \(\sqrt{1 - cos^{2} 36°}\)

= \(\sqrt{1 - (\frac{\sqrt{5} + 1}{4})^{2}}\)

= ¼\(\sqrt{10 - 2\sqrt{5}}\)

Therefore, a/b = \(\frac{\frac{1}{4}\sqrt{10 + 2\sqrt{5}}}{\frac{1}{4}\sqrt{10 - 2\sqrt{5}}}\)

= \(\frac{\sqrt{10 + 2\sqrt{5}}}{\sqrt{10 - 2\sqrt{5}}}\)

= \(\sqrt{\frac{(10 + 2\sqrt{5})^{2}}{10^{2} - (2\sqrt{5})^{2}}}\)

= \(\frac{10 + 2\sqrt{5}}{\sqrt{80}}\)

⇒ \(\frac{a}{b}\) = \(\frac{2√5(√5 + 1)}{4 √5}\)

⇒ \(\frac{a}{b}\) = \(\frac{√5 + 1}{2}\)

Therefore, a : b = (√5 + 1) : 2

**The Law of Sines or The Sine Rule****Theorem on Properties of Triangle****Projection Formulae****Proof of Projection Formulae****The Law of Cosines or The Cosine Rule****Area of a Triangle****Law of Tangents****Properties of Triangle Formulae****Problems on Properties of Triangle**

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