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JEE Main Syllabus 2025 (Physics, Chemistry, Maths)- Download PDF Here

Law of Cosines - Practice Questions & MCQ

Edited By admin | Updated on Sep 18, 2023 18:34 AM | #JEE Main

Quick Facts

  • Cosine Rule is considered one of the most asked concept.

  • 18 Questions around this concept.

Solve by difficulty

 A bird is sitting on the top of a vertical   pole 20 m high and its elevation from a point O on the ground is 450.  It flies off horizontally straight away from the point O.After one second, the elevation of the bird from O is reduced to 300. Then the speed (in m/s) of the bird is :

The angle of elevation of the top of a vertical tower from a point A, due east of it is 450.  The angle of elevation of the top of the same tower from a point B, due south of A is 300.  If the distance between A and B is 54\sqrt{2}m , then the height of the tower (in metres), is :

 Two ships A and B are sailing straight away from a fixed point O along routes such that \angleAOB is always 1200. At a certain instance, OA = 8 km, OB = 6 km and the ship A is sailing at the rate of 20 km/hr while the ship B sailing at the rate of 30 km/hr. Then the distance between A and B is changing at the rate (in km/hr) :

 If the angles of elevation of the top of a  tower from three collinear points A, B and C, on a line leading to the foot of the

tower, are 300, 450 and 600 respectively,then the ratio, AB : BC, is :

Concepts Covered - 1

Cosine Rule

Cosine Rule

For a triangle with angles A, B, and C, and opposite corresponding sides a, b, and c, respectively, the Law of Cosines is given as three equations. 

 

\cos A=\frac{b^{2}+c^{2}-a^{2}}{2 b c} \quad \cos B=\frac{a^{2}+c^{2}-b^{2}}{2 a c} \quad \cos C=\frac{a^{2}+b^{2}-c^{2}}{2 a b}

Proof:

Drop a perpendicular from C to the x-axis (this is the altitude or height). Recalling the basic trigonometric identities, we know that

\cos \theta=\frac{x(\text { adjacent })}{b(\text { hypotenuse })} \text { and } \sin \theta=\frac{y(\text { opposite })}{b(\text { hypotenuse })}

In terms of θ, x = bcos θ and y = bsin θ. The (x, y) point located at C has coordinates (bcos θ, bsin θ).

Using the side (x − c) as one leg of a right triangle and y as the second leg, we can find the length of hypotenuse a using the Pythagoras Theorem. Thus, 

\begin{aligned} a^{2} &=(x-c)^{2}+y^{2} \\ &=(b \cos \theta-c)^{2}+(b \sin \theta)^{2} \\&\quad\quad [\text { Substitute }(b \cos \theta) \text { for } x \text { and }(b \sin \theta) \text { for } y]\\ &=\left(b^{2} \cos ^{2} \theta-2 b c \cos \theta+c^{2}\right)+b^{2} \sin ^{2} \theta \\& \quad\quad[\text{Expand the perfect square.}] \\ &=b^{2} \cos ^{2} \theta+b^{2} \sin ^{2} \theta+c^{2}-2 bc \cos \theta \\&\quad \quad[\text { Group terms noting that } \cos ^{2} \theta+\sin ^{2} \theta=1]\\ &=b^{2}\left(\cos ^{2} \theta+\sin ^{2} \theta\right)+c^{2}-2 b c \cos \theta \\&\quad\quad[\text { Factor out } b^{2}] \\a^{2} &=b^{2}+c^{2}-2 b c \cos \theta \end{aligned}

Now as \theta equals angle A

\\a^2 = b^2 + c^2 - 2bc.cosA\\\\ cosA = \frac{b^2+c^2-a^2}{2bc}

Similarly we can derice formulae for cosB and cosC

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Cosine Rule

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