The length of AB joining Aa cos-0 and B0-a sin-0 is- a - 0

The correct option is A a Length =√((acosθ 0)^2+(0 sinθ)^2) =√(a^2cos^2θ+sin^2θ)=√(a)=a Hence length =a . ...

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Mathematics

Trigonometric Ratios of Standard Angles

The length of...

Question

The length of $¯ ¯¯¯¯¯¯ ¯ A B$ joining $A (a cos θ, 0)$ and $B (0, a sin θ, 0)$ is......... (a > 0)

a

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a^{2}

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a - 1

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Solution

The correct option is A $a$
Length $= \sqrt{(a cos θ - 0)^{2} + (0 - sin θ)^{2}}$
$= \sqrt{a^{2} {cos}^{2} θ + {sin}^{2} θ} = \sqrt{a} = a$
Hence length $= a$ .

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The length of ¯¯¯¯¯¯¯¯AB joining A(acosθ,0) and B(0,asinθ,0) is......... (a > 0)

The correct option is A aLength=√(acosθ−0)2+(0−sinθ)2=√a2cos2θ+sin2θ=√a=aHence length=a.

The length of $¯ ¯¯¯¯¯¯ ¯ A B$ joining $A (a cos θ, 0)$ and $B (0, a sin θ, 0)$ is......... (a > 0)

The correct option is A $a$
Length $= \sqrt{(a cos θ - 0)^{2} + (0 - sin θ)^{2}}$
$= \sqrt{a^{2} {cos}^{2} θ + {sin}^{2} θ} = \sqrt{a} = a$
Hence length $= a$ .