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Re: HOT Competition 4 Sep/8AM: Figure NOT drawn to scale. In the figure a [#permalink]
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if the degree measure of an inscribed angle subtending the red arc is 100° = angle at center
and the degree measure of an inscribed angle subtending the blue arc is 210 = angle at centre

y+ 105+50= 180
y= 25


180-x+50+105= 180
x= 155


x-y= 130

hence b
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Originally posted by mSKR on 04 Sep 2020, 08:56.
Last edited by mSKR on 05 Sep 2020, 08:47, edited 1 time in total.
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Re: HOT Competition 4 Sep/8AM: Figure NOT drawn to scale. In the figure a [#permalink]
1
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Refer to the attached diagram (screenshot)

Given

1. Subtended angle by Arc (Red) = 100 degrees
2. Subtended angle by Arc (Blue) = 210 degrees

We know that

Angle subtended by an arc at the centre of a circle = 2 * (Angle subtended by the same arc at some other

point on the circumference)


Using this property,

(1) In triangle BPD,

Angle (PBD) = 1/2 * Angle (COD)

= 1/2 * 210

Angle (PBD) = 105 degrees

Now, since an arc subtends equal angle in the same segment of a circle, therefore

Angle (PAC) = Angle (PBD) = 105 degrees


(2) In triangle APC,

Angle (PCA) = 1/2 * Angle (AOB)

= 1/2 * 100

Angle (PCA) = 50 degrees


Now, since an arc subtends equal angle in the same segment of a circle, therefore

Angle (PDB)
= Angle (PCA) = 50 degrees

(3) In triangles BPD,

y + 105 + 50 = 180

y = 25 degrees

Now, on a straight-line BC,

x + y = 180, so

x = 180 -y

x = 180 -25

x = 155 degrees


Therefore,

x – y = 155 – 25

x – y = 130 degrees

Option B (correct)

Posted from my mobile device
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IMG_20200905_190841.JPG [ 2.36 MiB | Viewed 2825 times ]

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Re: HOT Competition 4 Sep/8AM: Figure NOT drawn to scale. In the figure a [#permalink]
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Re: HOT Competition 4 Sep/8AM: Figure NOT drawn to scale. In the figure a [#permalink]
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