Given:
The figure.
To find:
The value of x.
Solution:
In triangle HIJ, HJ is diameter of the circle.
[tex]\angle HIJ=90^\circ[/tex] [Thales's Theorem]
[tex]HI=IJ[/tex] [Given]
[tex]\angle IHJ=\angle IJH=x-42^\circ[/tex] [Base angles of isosceles triangles are equal]
Now,
[tex]\angle HIJ+\angle IJH+\angle IHJ=180^\circ[/tex] [Angle sum property]
[tex](90^\circ)+(x-42^\circ)+(x-42^\circ)=180^\circ[/tex]
[tex]2x+6^\circ=180^\circ[/tex]
[tex]2x=180^\circ-6^\circ[/tex]
[tex]2x=174^\circ[/tex]
Divide both sides by 2.
[tex]x=\dfrac{174^\circ}{2}[/tex]
[tex]x=87^\circ[/tex]
Therefore, the value of x is [tex]87^\circ[/tex].
