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

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

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

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

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

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

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. |

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?

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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