iGCSE Mathematics (0580) :E4.6 Recognise rotaional and line symmetry.iGCSE Style Questions Paper 4

(a) Theorem: angle sum property of a triangle 

sum of all interior angles of a triangle =180

Therefore, \(\angle ATO\)+\(\angle AOT\)+\(\angle OAT\)=\(180^\circ\)

Here, \(\angle OAT\)=\(90^\circ\)

Given, \(\angle ATO\)=\(24^\circ\)

24+x+90=180

x=\(66^\circ\)

(b)  Here OB=OC (radius)

This implies that \(\angle OCB\)=\(\angle OBC\)(angles corresponding to equal are equal)

Let \(\angle OCB\)=\(\angle OBC\)=x

Now, TA is tangent 

This implies that \(\angle AOT\)=\(90^\circ\)(tangent is perpendicular to radius at the point of contact)

In triangle AOT, 

\(\angle OTA\)+\(\angle AOT\)+\(\angle OAT\)=\(180^\circ\)

90+24+\(\angle AOT\)=\(180^\circ\)

114+\(\angle AOT\)=\(180^\circ\)

\(\angle AOT\)=\(66^\circ\)

Now AC is diameter 

This implies that \(\angle AOT\)+\(\angle BOC\)=\(180^\circ\)(linear pair)

\(\angle BOC\)=180-66

\(\angle BOC\)=\(114^\circ\)

Now,\(\angle BOC\)+\(\angle OCB\)+\(\angle OBC\)=\(180^\circ\)

114+2x=180

x=33

\(\angle ACB\)=\(33^\circ\)

(c) In triangle AOB , AO=BO(radius)

\(\angle ABO\)=\(\angle BAO\)(angle corresponding to equal sides are equal)

Let \(\angle ABO\)=\(\angle BAO\)=y

\(\angle AOB\)+\(\angle OBA\)+\(\angle OAB\)=\(180^\circ\)(angle sum property of triangle )

66+2y=180

y=57

In triangle ABT ,

\(\angle AOB\)+\(\angle TAB\)=\(90^\circ\)

\(\angle TAB\)=90-54

\(\angle TAB\)=33

\(\angle ABT\)+\(\angle BAT\)\(\angle ATB\)=\(180^\circ\)

\(\angle ABT\)=\(123^\circ\)