
In the diagram, \[CB\] and \[CD\] are tangents to the circle with centre \[O\] . \[AOC\] is a straight line and \[\angle OCB = {34^ \circ }.\angle ABO\] equals.
A. \[{56^ \circ }\]
B. \[{28^ \circ }\]
C. \[{34^ \circ }\]
D. \[{32^ \circ }\]
Answer
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Hint: In this problem, we need to find the tangent of a circle with the given geometric representation line that touches the circle at a single point is known as a tangent to a circle. The point where tangent meets the circle is called the point of tangency.
Complete step-by-step answer:
In the given problem,
\[CB\] and \[CD\] are tangents to the circle with centre \[O\] .
\[AOC\] is a straight line and \[\angle OCB = {34^ \circ }.\angle ABO\] equals.
In \[\Delta AOB\] (Isosceles triangle property)
\[OA = OB\] (radius of the circle)
Here, a triangle whose two sides are equal and one side are unequal.
Therefore, \[\angle OBA = \angle OAB = x\] (Isosceles triangle property)
Then, the angle between the radius of the tangent
Also, \[\angle OBC = {90^ \circ }\]
Now,In \[\Delta AOC\] (Isosceles triangle property)
Sum of the property of the triangle is equal to \[{180^ \circ }\] .
\[\angle ACB + \angle BAC + \angle ABC = {180^ \circ }\]
According to the isosceles triangle whose two equal sides,one side is unequal.
\[2x + {90^ \circ } + {34^ \circ } = {180^ \circ }\]
By simplify it, we get
\[2x + {124^ \circ } = {180^ \circ }\]
Now, we get
\[
2x = {180^ \circ } - {124^ \circ } \\
x = \dfrac{{{{56}^ \circ }}}{2} = {28^ \circ } \\
\]
Hence, the isosceles triangle, \[\angle ABO = {28^ \circ }\]
The final answer is option(B) \[{28^ \circ }\]
So, the correct answer is “Option B”.
Note: In this question, tangent of the circle and isosceles triangle property are used to find the solution with respect to the problem.here, the tangent to a circle is defined as a straight line that touches the circle at a single point.The point where the tangent touches a circle and also isosceles triangle is the concept of two sides are equal and one unequal sides. These are the concepts used for finding the tangent of the circle.
Complete step-by-step answer:
In the given problem,
\[CB\] and \[CD\] are tangents to the circle with centre \[O\] .
\[AOC\] is a straight line and \[\angle OCB = {34^ \circ }.\angle ABO\] equals.
In \[\Delta AOB\] (Isosceles triangle property)
\[OA = OB\] (radius of the circle)
Here, a triangle whose two sides are equal and one side are unequal.
Therefore, \[\angle OBA = \angle OAB = x\] (Isosceles triangle property)
Then, the angle between the radius of the tangent
Also, \[\angle OBC = {90^ \circ }\]
Now,In \[\Delta AOC\] (Isosceles triangle property)
Sum of the property of the triangle is equal to \[{180^ \circ }\] .
\[\angle ACB + \angle BAC + \angle ABC = {180^ \circ }\]
According to the isosceles triangle whose two equal sides,one side is unequal.
\[2x + {90^ \circ } + {34^ \circ } = {180^ \circ }\]
By simplify it, we get
\[2x + {124^ \circ } = {180^ \circ }\]
Now, we get
\[
2x = {180^ \circ } - {124^ \circ } \\
x = \dfrac{{{{56}^ \circ }}}{2} = {28^ \circ } \\
\]
Hence, the isosceles triangle, \[\angle ABO = {28^ \circ }\]
The final answer is option(B) \[{28^ \circ }\]
So, the correct answer is “Option B”.
Note: In this question, tangent of the circle and isosceles triangle property are used to find the solution with respect to the problem.here, the tangent to a circle is defined as a straight line that touches the circle at a single point.The point where the tangent touches a circle and also isosceles triangle is the concept of two sides are equal and one unequal sides. These are the concepts used for finding the tangent of the circle.
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