# Law of Sines

Law of sines is, in triangle ABC :OR

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

In any triangle the three sides and the three angles are generally called the elements of the triangle. A triangle which does not contain a right angle is called

**an oblique triangle.**

In any triangle ABC, the measure of the angles $\angle$BAC, $\angle$CBA and $\angle$ACB are denoted by the letters A, B and C respectively. The sides BC, CA and AB opposite to the angles A, B and C respectively is denoted by a, b and c. These six elements of a triangle are not independent and are connected by the relations :

(i) A + B + C = π (180) (ii) a + b >c (iii) b + c > a (iv) c + a > b.

The elements of a triangle are connected by some trigonometric relations.

**Note : Law of sines is used when you know ASS and AAS**

**The LAW OF SINES OR SINE RULE**

**Theorem :**The side of an triangle are proportional to the sines of the angles opposite to them i.e. in a ΔABC. $\frac{a}{sin A}=\frac{b}{sin B}=\frac{c}{sin C}$

**Proof :**As the triangle may be acute, right and obtuse so there are three cases.

**Case I :**When ΔABC is an acute triangle.

When Δ ABC is an acute triangle. Draw AD perpendicular from A to the opposite side BC meeting it in the point D. In ΔABD, we have , sin B =$\frac{AD}{AB}$ => sin B = $\frac{AD}{c}$ =>AD = c . sin B ------(1) In ΔACD, we have , sin C =$\frac{AD}{AC}$ => sin C = $\frac{AD}{b}$ => AD = b . sin C ------(2) From equations (1) and (2) , we get c . sin B = b . sin C => $\frac{b}{sin B} = \frac{c}{sin C}$ Similarly, by drawing a perpendicular from B on AC, we have $\frac{a}{sin A} = \frac{c}{sin C}$ Hence, $\frac{a}{sin A}=\frac{b}{sin B} = \frac{c}{sin C}$ |

Case II : When Δ ABC is an obtuse triangle.Draw AD perpendicular from A to BC meet at D outside the triangle. In Δ ABD, we have, sin $\angle$ABD = $\frac{AD}{AB}$ => sin(180 - B) = $\frac{AD}{c}$ => AD = c.sin B -----(1) In Δ ACD, we have, sin C = $\frac{AD}{AC}$ => sin C = $\frac{AD}{b}$ => AD = b.sin C -----(2) From (1) and (2), we get, c.sin B = b.sin C => $\frac{b}{sin B} = \frac{c}{sin C}$ Similarly by drawing perpendicular fro B to AC, we have, $\frac{a}{sin A} = \frac{c}{sin C}$ Hence, $\frac{a}{sin A} = \frac{b}{sin B}=\frac{c}{sin C}$ |

Case III : When Δ ABC is a right triangle.In Δ ABC, we have sin C = sin (π/2) = 1. sin A =$ \frac{BC}{AB} = \frac{a}{c}$ and sin B =$ \frac{AC}{AB} = \frac{b}{c}$ &ther4; $\frac{a}{sin A} = \frac{b}{sin B}$ = c => $\frac{a}{sin A} =\frac{b}{sin B} = \frac{c}{1}$ => $\frac{a}{sin A} =\frac{b}{sin B} = \frac{c}{sin C}$ (since sin C = sin (π/2) = 1) Hence in all cases, we get $\frac{a}{sin A} =\frac{b}{sin B} = \frac{c}{sin C}$ |

## Examples on law of sines

1)Two surveyors are determining the distance to a tower located between them but across the river. The first one determines that the line of sight to the tower makes an angle of 89° with the bank of the river. 495 m downstream another surveyor determines his line of sight to the tower is 55° to the tower with the river. How far is each surveyor to the tower?**Solution :**According to the diagram the measure of 3rd angle will be $36^{0}$.

To find the value of x, by the law of sines,

$\frac{sin 36}{495}= \frac{sin 55}{x}$

$\frac{0.58778}{495}= \frac{0.81915}{x}$

$x = \frac{0.81915 \times 495}{0.58778}$ (by cross multiplication)

∴ x = 689.84 km.

To find the value of y, by the law of sines,

$\frac{sin 36}{495}= \frac{sin 89}{y}$

$\frac{0.58778}{495}= \frac{0.99984}{y}$

$y = \frac{0.99984 \times 495}{0.58778}$ (by cross multiplication)

∴ y = 842.02 km.

2) Two points M and N are separated by a swamp. A base line MK is established on one side of the swamp. MK is 180 m in length. The measure of angle K is 62° and angle N is 74°. Find the distance between M and N.

**Solution :**According to the diagram, let the length of the swamp be x.

To find the value of x, by the law of sines,

$\frac{sin 62}{x}= \frac{sin 74}{180}$

$\frac{0.88294}{x}= \frac{0.96126}{180}$

$x = \frac{0.88294 \times 180}{0.96216}$ (by cross multiplication)

∴ x = 165.18 km.

precalculus

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