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# The distance between the points (a cos 25°, 0) and (0, a cos 65°) is - Mathematics

MCQ

The distance between the points (a cos 25°, 0) and (0, a cos 65°) is

#### Options

• a

• 2a

• 3a

•  None of these

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#### Solution

We have to find the distance between A(a cos 25°, 0)  and B (0 , a cos 65° ) .

In general, the distance between A(x1 , y1 )  and B( x2 ,y2 ) is given by,

AB = sqrt((x_2 - x_1)^2 + (y_2 - y_1)^2)

So,

$AB = \sqrt{\left( 0 - a\cos25° \right)^2 + \left( a\cos65° - 0 \right)^2}$
$= \sqrt{\left( a\cos25° \right)^2 + \left( a\cos65° \right)^2}$

$\cos25° = \sin65° and \cos65° = \sin25°$

But according to the trigonometric identity,

sin^2  theta + cos^2 theta = 1

Therefore,

AB = a

Is there an error in this question or solution?
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#### APPEARS IN

RD Sharma Class 10 Maths
Chapter 6 Co-Ordinate Geometry
Q 2 | Page 63
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