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Law of CosinesLaw 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
Examples on law of cosines1)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 Home Covid19 has led the world to go through a phenomenal transition . Elearning is the future today. Stay Home , Stay Safe and keep learning!!! Covid19 has affected physical interactions between people. Don't let it affect your learning.
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