# Law of Cosines

Law of cosines : In a non right triangle,when all the sides are given and students have to find the measure of angle then students can use law of cosines. This cosine rule is also applicable to determined the 3rd side and when two sides and one angle between them is given.**Theorem :**In a triangle ABC,

1) $a^{2}=b^{2} + c^{2} -2bc.CosA$

2) $b{2}=a^{2} + c^{2} -2ac.CosB$

3) $c^{2}=a^{2} + b^{2} -2ab.CosC$

**Formulas to find the angle between the two sides :**

1) $cos A = \frac{b^{2}+ c^{2} - a^{2}}{2bc}$

2) $cos B = \frac{c^{2}+ a^{2} - b^{2}}{2ac} $

3) $cos C = \frac{a^{2}+ b^{2} - c^{2}}{2ab}$

**Note : Law of cosines is used when you know SSS and SAS**

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 , Cos B = $\frac{BD}{c}$ => BD = c. CosB ------- (1) In ΔACD, we have Cos C = $\frac{CD}{b}$ => CD = b. CosC In ΔACD, by Pythagorean theorem, we have, $ AC^{2} = AD^{2} + CD^{2}$ => $AC^{2} = AD^{2} + (BC - BD)^{2}$ => $AC^{2} = AD^{2} + BC^{2} + BD^{2}- 2BC.BD$ => $AC^{2} = BC^{2} + (AD^{2} +BD^{2})- 2BC.BD$ =>$ AC^{2} = BC^{2} + AB^{2}- 2BC.BD$ ----(2) (since $AD^{2} +BD^{2} = AB^{2}$) =>$ b^{2} = a^{2} + c^{2}- 2ac.CosB $ [ by equation (1)] => $cos B = \frac{c^{2}+ a^{2} - b^{2}}{2ac} $ |

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, Cos(180 - B) = $\frac{BD}{AB}$ => BD = - AB cos B => - c CosB------(3) In ΔACD, by Pythagorean theorem, we have, $ AC^{2} = BC^{2} + AB^{2}- 2BC.BD$ [ from equation (2)] $ AC^{2} = AD^{2} + CD^{2}$ => $AC^{2} = AD^{2} + (BC + BD)^{2}$ => $AC^{2} = AD^{2} + BC^{2} + BD^{2}+ 2BC.BD$ => $AC^{2} = BC^{2} + (AD^{2} +BD^{2})+ 2BC.BD$ =>$ AC^{2} = BC^{2} + AB^{2}+ 2BC.BD$ (since $AD^{2} +BD^{2} = AB^{2}$) =>$ b^{2} = a^{2} + c^{2}- 2a(-c.CosB) $ [ by equation (3)] =>$ b^{2} = a^{2} + c^{2}- 2ac.CosB $ => $cos B = \frac{c^{2}+ a^{2} - b^{2}}{2ac} $ |

Case III: When Δ ABC is right triangle. Let Δ ABC, be a right angle at B, then by Pythagorean theorem. $b^{2} = a^{2} + c^{2}$ => $b^{2} = a^{2} + c^{2} - 2 ac cosB $ ( since angle B = $\frac{\pi }{2}$ => CosB = 0) => $cos B = \frac{c^{2}+ a^{2} - b^{2}}{2ac} $ Hence in all the cases , we have $b^{2} = a^{2} + c^{2} - 2 ac. cosB $ Similarly, other cosine rules or other results can be proved. |

## Examples on law of cosines

1)Two cruises leave the same port. Cruise A travels at 30 km/h. Cruise B travels at 37 km/h. The angle between their path is $62^{0}$. How far are the two cruises apart three hours later?**Solution :**In the triangle APB, two sides and angle between them is given so we will use a cosine rule here.

Rate of Cruise A = 30km/hr

So distance traveled by Cruise A = 30 x 3 = 90 km.

Rate of Cruise B = 37 km/hr

So distance traveled by Cruise B = 37 x 3 = 111 km.

The angle APB= $62^{0}$.

By cosine rule,

$AB^{2}= 90^{2} +111^{2} -2(90)(111).cos(62)$

= 8100 + 12321 - 19980(0.46947)

= 20421 - 9380.0106

$AB^{2}$ = 11040.9894

∴ AB = 105.07 km.

2) A hot air balloon B is fixed to the ground at F and G by 2 ropes 120m and 150 m long. If angle FBG is $86^{0}$.How far apart are F and G?

**Solution :**In the triangle FBG, two sides and angle between them is given so we will use a cosine rule here.

Rope FB = 150 m

Rope BG = 120 m

Angle FBG= $86^{0}$ By cosine rule,

$FG^{2}= 150^{2} + 120^{2} -2(150)(120).cos(86)$

= 22500 + 14400 - 36000(0.0697564)

= 36900 - 2511.2304

$FG^{2}$ = 34388.7696

∴ FG = 185.4 m.

precalculus

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