Answer:
by exterior angle property.
x=45°
y=15°
[tex]\large\underline{\sf{Solution-}}[/tex]
Given that,
∠AOD = 90°
and
∠AEC = 90°
But these are corresponding angles.
[tex]\implies \: [/tex] OD || BC
So, ∠OCB = ∠DOC = 30° [Alternate interior angles]
Now, we know, angle subtended at the centre of a circle by an arc is double the angle subtended on the circumference of a circle by same arc.
[tex] \implies \:[/tex] ∠COD = 2∠CBD
[tex]\rm\implies \:2y = 30 \degree \: \\ [/tex]
[tex]\bf\implies \:y = 15\degree \: \\ \\ [/tex]
Again, angle subtended at the centre of a circle by an arc is double the angle subtended on the circumference of a circle by same arc.
[tex] \rm\implies \:[/tex]∠AOD = 2∠ABD
[tex]\rm\implies \:2z = 90 \degree \: \\ [/tex]
[tex]\bf\implies \:z = 45 \degree \: \\ \\ [/tex]
Now, In triangle ABE,
We know, Exterior angle of a triangle is equals to sum of interior opposite angles.
[tex]\rm\implies \: [/tex]∠AEC = ∠BAE + ∠ABE
[tex]\rm\implies \:x + (y + z) = 90 \degree \\ [/tex]
[tex]\rm\implies \:x + (15\degree + 45\degree) = 90 \degree \\ [/tex]
[tex]\rm\implies \:x + 60\degree = 90 \degree \\ [/tex]
[tex]\bf\implies \:x = 30 \degree \\ \\ [/tex]
[tex]\rule{190pt}{2pt} \\ [/tex]
[tex] { \red{ \mathfrak{Additional\:Information}}}[/tex]
1. Angle in same segments are equal.
2. Angle in semi-circle is 90°.
3. Sum of opposite angles of a cyclic quadrilateral is 180°.
4. Equal chord subtends equal angles at the center.
5. Equal chords are equidistant from center.
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Answers & Comments
Answer:
by exterior angle property.
x=45°
y=15°
Verified answer
[tex]\large\underline{\sf{Solution-}}[/tex]
Given that,
∠AOD = 90°
and
∠AEC = 90°
But these are corresponding angles.
[tex]\implies \: [/tex] OD || BC
So, ∠OCB = ∠DOC = 30° [Alternate interior angles]
Now, we know, angle subtended at the centre of a circle by an arc is double the angle subtended on the circumference of a circle by same arc.
[tex] \implies \:[/tex] ∠COD = 2∠CBD
[tex]\rm\implies \:2y = 30 \degree \: \\ [/tex]
[tex]\bf\implies \:y = 15\degree \: \\ \\ [/tex]
Again, angle subtended at the centre of a circle by an arc is double the angle subtended on the circumference of a circle by same arc.
[tex] \rm\implies \:[/tex]∠AOD = 2∠ABD
[tex]\rm\implies \:2z = 90 \degree \: \\ [/tex]
[tex]\bf\implies \:z = 45 \degree \: \\ \\ [/tex]
Now, In triangle ABE,
We know, Exterior angle of a triangle is equals to sum of interior opposite angles.
[tex]\rm\implies \: [/tex]∠AEC = ∠BAE + ∠ABE
[tex]\rm\implies \:x + (y + z) = 90 \degree \\ [/tex]
[tex]\rm\implies \:x + (15\degree + 45\degree) = 90 \degree \\ [/tex]
[tex]\rm\implies \:x + 60\degree = 90 \degree \\ [/tex]
[tex]\bf\implies \:x = 30 \degree \\ \\ [/tex]
[tex]\rule{190pt}{2pt} \\ [/tex]
[tex] { \red{ \mathfrak{Additional\:Information}}}[/tex]
1. Angle in same segments are equal.
2. Angle in semi-circle is 90°.
3. Sum of opposite angles of a cyclic quadrilateral is 180°.
4. Equal chord subtends equal angles at the center.
5. Equal chords are equidistant from center.