If the length of a chord of a circle, which makes an angle 45° with the tangent drawn at one end point of the chord, is 6cm, then the radius of the circle is :
Let the chord AB = 6 .'M' be it's midpoint . 'O' be the centre. The tangent MN meets at B and OB makes 90° . Given that angle by chord at B = 45°.
So, angle OAM = 45°= angle BOM. Also ,MB = 3 .
$$\frac{OB}{sin90}=\frac{3}{sin45}$$
we get OB= 3√2
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