If the line segment joining the points A a - b and B c - d subtends an angle - at the origin- then cos- is equal toA- a b-c d-a2-b2c2-d2B- a c-b d-a2-b2c2-d2C- a c-b d-a2-b2c2-d2D- None of these

The correct option is B ac+bd/√((a^2+b^2)(c^2+d^2)) Here (AB)^2= (a c)^2+(b d)^2 (OA)^2=(a 0)^2+(b 0)^2=a^2+b^2 and (OB)^2=c^2+d^2 In Δ AOB, cosθ= (OA)^2+ ...

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Standard XII

Mathematics

Cosine Rule

If the line s...

Question

If the line segment joining the points $A (a, b)$ and $B (c, d)$ subtends an angle $θ$ at the origin, then $cos θ$ is equal to

$\frac{a b + c d}{\sqrt{(a^{2} + b^{2}) (c^{2} + d^{2})}}$

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$\frac{a c + b d}{\sqrt{(a^{2} + b^{2}) (c^{2} + d^{2})}}$

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$\frac{a c - b d}{\sqrt{(a^{2} + b^{2}) (c^{2} + d^{2})}}$

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None of these

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Solution

The correct option is B
$\frac{a c + b d}{\sqrt{(a^{2} + b^{2}) (c^{2} + d^{2})}}$

Here $(A B)^{2} = (a - c)^{2} + (b - d)^{2} (O A)^{2} = (a - 0)^{2} + (b - 0)^{2} = a^{2} + b^{2} and (O B)^{2} = c^{2} + d^{2}$

$In Δ A O B, cos θ = \frac{(O A)^{2} + (O B)^{2} - (A B)^{2}}{2 O A . O B} = \frac{a^{2} + b^{2} + c^{2} + d^{2} - (a - c)^{2} + (b - d)^{2}}{2 \sqrt{a^{2} + b^{2}} . \sqrt{c^{2} + d^{2}}} \frac{a c + b d}{\sqrt{(a^{2} + b^{2}) (c^{2} + d^{2})}} .$

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Similar questions

Q. If the segments joining the points

A (a, b)

and

B (c, d)

studends an angle at the origin, prove that

cos θ = \frac{a c + b d}{\sqrt{(a^{2} + b^{2}) (c^{2} + d^{2})}}

Q. If the line segment joining the points

A (a, b)

and

B (c, d)

subtends an angle

θ

at the origin,then

cos θ = a c + b d \sqrt{(a^{2} + b^{2}) (c^{2} + d^{2})}

Factorize $(a^{2} - b^{2}) (c^{2} - d^{2}) - 4 a b c d$

Q. If the line segment joining the points

A (a, b)

and

B (c, d)

subtends an angle

θ

at the origin, then

cos θ

is equal to

Q. Question 4
If

s i n θ = \frac{a}{b}

, then

c o s θ

is equal to

(A)

\frac{b}{\sqrt{b^{2} - a^{2}}}

(B)

\frac{b}{a}

(C)

\frac{\sqrt{b^{2} - a^{2}}}{b}

(D)

\frac{a}{\sqrt{b^{2} - a^{2}}}

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If the line segment joining the points A(a,b) and B(c,d) subtends an angle θ at the origin, then cosθ is equal to

The correct option is B ac+bd√(a2+b2)(c2+d2) Here (AB)2=(a−c)2+(b−d)2(OA)2=(a−0)2+(b−0)2=a2+b2 and (OB)2=c2+d2 In ΔAOB, cosθ=(OA)2+(OB)2−(AB)22OA.OB=a2+b2+c2+d2−(a−c)2+(b−d)22√a2+b2.√c2+d2ac+bd√(a2+b2)(c2+d2).

If the line segment joining the points $A (a, b)$ and $B (c, d)$ subtends an angle $θ$ at the origin, then $cos θ$ is equal to