Cosine Rule MCQ Quiz in मल्याळम - Objective Question with Answer for Cosine Rule - സൗജന്യ PDF ഡൗൺലോഡ് ചെയ്യുക

Given:

Two chords of length 'a' unit and 'b' unit of a circle make angles 60° and 90° at the centre of a circle respectively,

The formula used:

\(\text {cos} (θ) = \frac{a^2 + b^2 - c^2}{2ab}\)

Where, 

\(a, b\) and \(c\) are the side of the triangle and \(\theta\) is the angle opposite to the side \(c\).

The Pythagoras Theorem:

\(h^2 = p^2 + b^2\)

Where, 

\(h, p \text { and } b\) are hypotenuse, perpendicular and base respectively.

Calculation:

According to the question, the required image is:

In Δ AOB,

The length of the chord or the base of the triangle AOB \(=a\)

Therefore, 

\(\Rightarrow \text {cos} (60^\circ) = \frac{r^2 + r^2 - a^2}{2r^2} \\ \Rightarrow \frac{1}{2} = \frac{2r^2 - a^2}{2r^2} \\ \Rightarrow 1 = \frac{2r^2 - a^2}{r^2} \\ \Rightarrow r^2 = a^2 \\ \Rightarrow a = r \,\,\,\,\,\,......(i)\)

In Δ DOC,

The length of the required chord or the base of the triangle is \(b\).

By using the Pythagoras theorem, we get,

\(\Rightarrow r^2 + r^2 = b^2 \\ \Rightarrow 2r^2 = b^2 \\ \Rightarrow b = √{2}r \,\,\,\,\,\,......(ii)\)

From equation \((i)\) and \((ii)\), we get,

\(\Rightarrow b=√{2}a\)

 ∴ The required relation is \(b=\sqrt{2}a\).

Concept:

Cosine Rule:

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

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

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

Calculation:

Given: c(a.cos B - b.cos A)

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

⇒ b.c.cos A = \(\rm \frac {b^2 + c^2 - a^2}{2}\)      ----(1)

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

⇒ a.c.cos B = \(\rm \frac {c^2 + a^2 - b^2}{2}\)      ----(2)

Subtract equation (2) from (1) we get

⇒ a.c.cos B - b.c.cos A = \(\rm (\frac {c^2 + a^2 - b^2}{2}) - ( \frac {b^2 + c^2 - a^2}{2})\)

⇒ c(a.cos B - b.cos A) = \(\rm \frac {({c^2 + a^2 - b^2 - b^2 - c^2 +a^2})}{2}\)

⇒ c(a.cos B - b.cos A) = \(\rm \frac {2({a^2 - b^2})}{2}\)

⇒ c(a.cos B - b.cos A) = a² - b²

Additional Information

b(c.cos A - a.cos C) = c² - a²

a(b.cos C - c.cos B) = b² - c²

Concept:

Cosine rule:

\(\rm \cos θ = {a^2+ b^2 - c^2\over2ab}\)

Where a, b and c are the sides of the triangle and θ is the angle between the sides a and b.

Calculation:

Given a = 5, b = 10 and θ = 60°

So according to the cosine rule,

\(\rm \cos 60= {5^2+10^2-c^2\over2(5)(10)}\)

⇒ \(\rm {1\over2}= {25\ +\ 100\ -\ c^2\over2\ \times \ 50}\)

⇒ 50 = 125 - c 2

⇒ c2 = 125 - 50

⇒ c² = 75

c = 5√2

Additional Information

Cosine rule:

\(\rm \cos C = {a^2+ b^2 - c^2\over2ab}\)

\(\rm \cos B = {a^2+ c^2 - b^2\over2ac}\)

\(\rm \cos A = {c^2+ b^2 - a^2\over2cb}\)

Sine rule:

\(\frac{{\rm{a}}}{{\sin {\rm{A}}}} = \frac{{\rm{b}}}{{\sin {\rm{B}}}} = \frac{{\rm{c}}}{{\sin {\rm{C}}}}\)