WAEC 2010 · Paper 2 · Q3✱✱

(a)
In the diagram, $PQ \parallel UT$ , $\angle PQR = 125^\circ$ , $\angle QRS = r$ , $\angle RST = 80^\circ$ and $\angle STU = 44^\circ$ . Calculate the value of $r$ .
PQ is parallel to UT. Not to scale.
(b)
In the diagram, $TS$ is a tangent to the circle at $A$ . $AB \parallel CE$ , $\angle AEC = 5x^\circ$ , $\angle ADB = 60^\circ$ and $\angle TAE = x^\circ$ . Find the value of $x$ .

Worked solution (try it first)

Draw a line through $R$ and a line through $S$ , both parallel to $PQ$ and $UT$ .
They split $r$ and $\angle RST$ into pieces you can find.
At $Q$ : the angle between $QR$ and the line through $R$ is $180^\circ - 125^\circ = 55^\circ$ (co-interior angles add up to $180^\circ$ ).
At $S$ : the lower piece of $\angle RST$ is $44^\circ$ (alternate to $\angle STU$ ).
So the upper piece of $\angle RST$ is $80^\circ - 44^\circ = 36^\circ$ , and the piece of $r$ below the line through $R$ is also $36^\circ$ (alternate angles).
Add the two pieces: $r = 55^\circ + 36^\circ = 91^\circ$ .

Angle in the alternate segment: $\angle BAS = \angle ADB = 60^\circ$ .
$AB \parallel CE$ , so $\angle BAE$ and $\angle AEC$ are co-interior: $\angle BAE = 180^\circ - 5x$ .
The angles at $A$ on the straight line $SAT$ add up to $180^\circ$ : $60 + (180 - 5x) + x = 180$ .
Simplify: $240 - 4x = 180$ , so $4x = 60$ and $x = 15$ .