Given: measure of angle KJ = 124°, measure of IC =38°
Find: m∠CQJ, m∠LIJ.
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Answer:
Part 1) [tex]m<CQJ=99\°[/tex]
Part 2) [tex]m<LIJ=118\°[/tex]
Step-by-step explanation:
Part 1) Find the measure of angle CQJ
we know that
The measure of the interior angle is the semi-sum of the arcs comprising it and its opposite.
[tex]m<CQJ=\frac{1}{2}(arc\ IK+arc\ CJ)[/tex]
[tex]arc\ IK+arc\ CJ=360\°-(arc\ IC+arc\ KJ)[/tex]
substitute the values
[tex]arc\ IK+arc\ CJ=360\°-(38\°+124\°)=198\°[/tex]
[tex]m<CQJ=\frac{1}{2}(198\°)=99\°[/tex]
Part 2) Find the measure of angle LIJ
step 1
Find the measure of angle IJL
we know that
The inscribed angle is half that of the arc it comprises.
[tex]m<IJL=\frac{1}{2}(arc\ IC)[/tex]
substitute the values
[tex]m<IJL=\frac{1}{2}(38\°)=19\°[/tex]
step 2
Find the measure of angle ILJ
we know that
The measurement of the external angle is the semi-difference of the arcs it encompasses.
[tex]m<ILJ=\frac{1}{2}(arc\ KJ-arc\ IC)[/tex]
substitute the values
[tex]m<ILJ=\frac{1}{2}(124\°-38\°)=43\°[/tex]
step 3
Find the measure of angle LIJ
Remember that the sum of the interior angles of a triangle must be equal to 180 degrees
In the triangle LIJ
[tex]m<LIJ+m<ILJ+m<IJL=180\°[/tex]
substitute the values
[tex]m<LIJ+43\°+19\°=180\°[/tex]
[tex]m<LIJ=180\°-62\°=118\°[/tex]
A circle is a curve sketched out by a point moving in a plane. The measurements of the angle ∠CQJ and ∠LIJ are 99° and 118° respectively.
A circle is a curve sketched out by a point moving in a plane so that its distance from a given point is constant; alternatively, it is the shape formed by all points in a plane that are at a set distance from a given point, the centre.
Given that measure of KJ is 124°, therefore, the measure of the ∠KOJ is 124°. Since the angle formed at the centre and at any point on the circle form the same chord are in the ratio of 2:1. Thus,
[tex]2\angle KCJ = \angle KOJ = 124^o\\\\\angle KCJ = 62^o[/tex]
Similarly, the measure of IC is 38°, therefore, the measure of ∠IOC is 38°. Further, it can be written as,
[tex]\angle IOC = 2\angle IJC = 38 ^o\\\\\angle IJC = 19^o[/tex]
Now, In the triangle CQJ, the sum of all the angles can be written as,
[tex]\angle CQJ + \angle QJC + \angle QCJ = 180^o\angle CQJ = 99^o[/tex]
As the ∠IKC and ∠IJC are formed from the same points in the circle their measurement will be equal as well, therefore,
[tex]\angle IKC = \angle IJC = 19^o[/tex]
Since ∠KCL and ∠KCj are supplementary angles, their measure can be written as,
[tex]\angle KCL + \angle KCJ = 180^o\\\angle KCL = 118^o[/tex]
In the triangle LCK, the sum of the angles can be written as,
[tex]\angle KCL+ \angle CKL+ \angle CLK= 180^o\\\\118^o + 19^o + \angle CLK= 180^o\\\\\angle CLK= 43^o[/tex]
In the triangle LIJ, the sum of the angles can be written as,
[tex]\angle LIJ+ \angle IJL+ \angle JLI= 180^o\\\\ \angle LIJ + 19^o + 43^o= 180^o\\\\\angle CLK= 118^o[/tex]
Thus, the measurements of the angle ∠CQJ and ∠LIJ are 99° and 118° respectively.
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