We will discuss here about the law of cosines or the cosine rule which is required for solving the problems on triangle.
In any triangle ABC, Prove that,
(i) b\(^{2}\) = c\(^{2}\) + a\(^{2}\) - 2ca. cos B or, cos B = \(\frac{c^{2} + a^{2} - b^{2}}{2ca}\)
(ii) a\(^{2}\) = b\(^{2}\) + c\(^{2}\) - 2ab. cos A or, cos A = \(\frac{b^{2} + c^{2} - a^{2}}{2bc}\)
(iii) c\(^{2}\) = a\(^{2}\) + b\(^{2}\) - 2ab. cos C or, cos C = \(\frac{a^{2} + b^{2} - c^{2}}{2ab}\)
Proof of the law of cosines:
Let ABC is a triangle. Then the following three cases arise:
Case I: When the triangle ABC is acute-angled:
Now form the triangle ABD, we have,
cos B = BD/BC
⇒ cos B = BD/c
⇒ BD = c cos B ……………………………………. (1)
Again from the triangle ACD, we have
cos C = CD/CA
⇒ cos C = CD/b
⇒ CD = b cos C
By using the Pythagoras theorem on the triangle ACD, we get
AC\(^{2}\) = AD\(^{2}\) + CD\(^{2}\)
⇒ AC\(^{2}\) = AD\(^{2}\) + (BC - BD)\(^{2}\)
⇒ AC\(^{2}\) = AD\(^{2}\) + BC\(^{2}\) + BD\(^{2}\) - 2 BC ∙ BD
⇒ AC\(^{2}\) = BC\(^{2}\) + (AD\(^{2}\) + BD\(^{2}\)) - 2 BC ∙ BD
⇒ AC\(^{2}\) = BC\(^{2}\) + AB\(^{2}\) - 2 BC ∙ BD, [Since From triangle, we get, AD\(^{2}\) + BD\(^{2}\) = AB\(^{2}\)]
⇒ b\(^{2}\) = a\(^{2}\) + c\(^{2}\) - 2a ∙ c cos B, [From (1)]
⇒ b\(^{2}\) = c\(^{2}\) + a\(^{2}\) - 2ca cos B or, cos B = \(\frac{c^{2} + a^{2} - b^{2}}{2ca}\)
Case II: When the triangle ABC is obtuse-angled:
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,
cos (180° - B) = BD/AB
⇒- cos B = BD/AB, [Since, cos (180° - B) = - cos B]
⇒ BD = -AB cos B
⇒ BD = -c cos B ……………………………………. (2)
By using the
Pythagoras theorem on the triangle ACD, we get
AC\(^{2}\) = AD\(^{2}\) + CD\(^{2}\)
⇒ AC\(^{2}\) = AD\(^{2}\) + (BC + BD)\(^{2}\)
⇒ AC\(^{2}\) = AD\(^{2}\) + BC\(^{2}\) + BD\(^{2}\) + 2 BC ∙ BD
⇒ AC\(^{2}\)= BC\(^{2}\)+ (AD^2 + BD^2) + 2 BC ∙ BD
⇒ AC\(^{2}\) = BC\(^{2}\) + AB\(^{2}\) + 2 BC ∙ BD, [Since From triangle, we get, AD\(^{2}\) + BD\(^{2}\) = AB\(^{2}\)]
⇒ b\(^{2}\) = a\(^{2}\) + c\(^{2}\) + 2a ∙ (-c - cos B), [From (2)]
⇒ b\(^{2}\) = c\(^{2}\) + a\(^{2}\) - 2ca cos B or, cos B = \(\frac{c^{2} + a^{2} - b^{2}}{2ca}\)
Case III: Right angled triangle (one angle is right angle): The triangle ABC is right angled. The angle B is a right angle.
Now by using the Pythagoras theorem we get,
b\(^{2}\) = AC\(^{2}\) = BC\(^{2}\) + BA\(^{2}\) = a\(^{2}\) + c\(^{2}\)
⇒ b\(^{2}\) = a\(^{2}\) + c\(^{2}\)
⇒ b\(^{2}\) = a\(^{2}\) + c\(^{2}\) - 2ac cos B, [We know that cos 90° = 0 and B = 90°. Therefore, cos B = 0] or, cos B = \(\frac{c^{2} + a^{2} - b^{2}}{2ca}\)
Therefore, in all three cases, we get,
b\(^{2}\) = a\(^{2}\) + c\(^{2}\) - 2ac cos B or, cos B = \(\frac{c^{2} + a^{2} - b^{2}}{2ca}\)
Similarly, we can prove that the formulae (ii) a\(^{2}\) = b\(^{2}\) + c\(^{2}\) - 2ab. cos A or, cos A = \(\frac{b^{2} + c^{2} - a^{2}}{2bc}\) and (iii) c\(^{2}\) = a\(^{2}\) + b\(^{2}\) - 2ab. cos C or, cos C = \(\frac{a^{2} + b^{2} - c^{2}}{2ab}\).
Solved problem using the law of Cosines:
In the triangle ABC, if a = 5, b = 7 and c = 3; find the angle B and the circum-radius R.
Solution:
Using the formula, cos B = \(\frac{c^{2} + a^{2} - b^{2}}{2ca}\) we get,
cos B = \(\frac{3^{2} + 5^{2} - 7^{2}}{2 ∙ 3 ∙ 5}\)
cos B = \(\frac{9 + 25 - 49}{30}\)
cos B = - 1/2
cos B = cos 120°
Therefore, B = 120°
Again, if R be the required circum-radius then,
b/sin B = 2R
⇒ 2R = 7/sin 120°
⇒ 2R = 7 ∙ 2/√3
Therefore, R = 7/√3 = (7√3)/3 units.
11 and 12 Grade Math
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