Cosine Rule MCQ Quiz in मल्याळम - Objective Question with Answer for Cosine Rule - സൗജന്യ PDF ഡൗൺലോഡ് ചെയ്യുക
Last updated on Mar 14, 2025
Latest Cosine Rule MCQ Objective Questions
Top Cosine Rule MCQ Objective Questions
Cosine Rule Question 1:
. Two chords of length a unit and b unit of a circle make angles 60° and 90° at the centre of a circle respectively, then the correct relation is:
Answer (Detailed Solution Below)
Cosine Rule Question 1 Detailed Solution
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 \(b=\sqrt{2}a\).
Cosine Rule Question 2:
For any ΔABC find the value of c(a⋅ cos B - b⋅ cos A)
Answer (Detailed Solution Below)
Cosine Rule Question 2 Detailed Solution
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
Cosine Rule Question 3:
The angle between the two sides of the triangle having lengths 5 cm and 10 cm is 60°, find the third side of the triangle.
Answer (Detailed Solution Below)
Cosine Rule Question 3 Detailed Solution
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:
\(\frac{{\rm{a}}}{{\sin {\rm{A}}}} = \frac{{\rm{b}}}{{\sin {\rm{B}}}} = \frac{{\rm{c}}}{{\sin {\rm{C}}}}\)