1)
a. Angle p.
b. Angle d.
c. Assuming that 'interior angle' means 'same side interior angle', angle p.
d. Angle p.
e. Angle c.
Angles c and r are corresponding angles because they're at the same location at points where a transversal intersects parallel lines. You can have infinite points on a line. So transversal can have lines that are parallel to one another intersect at every one of those points. So since the angles at which the parallel lines intersect the transversal will correspond for all the parallel lines, they'll all have the same value.
If you take question 2, then angles AQR, CRS and EST are also corresponding angles. Corresponding angles are congruent.
Hereon I'll leave the solutions, but omit the actual answer...
2)
a. Angles BQR and DRS are corresponding angles too, which means they're congruent—and congruent angles have the same values.
b. Angles BQR and QRC are alternate angles (aka opposite angles, as I learnt them). Opposite angles are also congruent.
c. The sum of values of all angles that make up a straight line is 180º, because the angle at any point on a straight line is 180º. Angles DRQ and QRC make up the straight line CD, so DRQ + QRC = 180º.
d. Again, angles RSE and QRC are corresponding angles.
e. Again, corresponding angles with angles BQR and DRS.
f. This is getting repetitive—RSF and DRQ and corresponding angles.
3)
a. AEF and CFE are interior angles (don't confuse with alternate interior angles). The sum of values of interior angles is 180º (it's an extension of the straight line principle that says the sum of angles that make up a straight line will be 180º). If that's confusing for you, think about it this way: CFH and CFE form a straight line, so 180º - CFH = CFE; and the CFE corresponds to AEF.
b. I'll go with the shorter method for this one. Here, AEF, BEG and the right angle FEG form a straight line. So, AEF + FEG + BEG = 180º. But we know the value of AEF, and right angles value at 90º, so it's simple arithmetic to find out BEG.
c. EGF and BEG are corresponding. Your workbook really loves corresponding angles...
d. We figured out CFH in 3a.—CFH and AEF are corresponding.
4) BF and CD are parallel line segments and the transversals are CA and DF.
a.
i. Angle F is an alternate angle (thank God for the variety) to angle FDE. Yet another pair of congruent angles.
ii. Angle C is correspondent with angle ABF over transversal CA. Will we ever forget now that correspondent angles are congruent?
b. For this you need angles F, C, CBF and CDF. You can use the straight line principle to find the values of the second pair of angles (e.g., CBF + ABF = 180º; => CBF = 180 - ABF = 180 - 75, and likewise with CDF). This will add up to 360º, and this is an important and useful principle: the sum of angles in any quadrilateral (any four sided shape, even irregular shapes) will always be 360º. Just like how a triangle's angles will always add up to 180º.
If you've already learnt that bit about a triangle's angles in class, then you can show the sum of angles for quadrilaterals in a neat way. Draw a diagonal in the trapezium (or as you Americans call it, trapezoid): this will cut it into two triangles. The sum of angles of a triangle is 180º—so the sum of angles in two triangles will be 180 + 180 = 360º.
5) Opposite sides are parallel in a parallelogram, so you want to keep using interior angles to find everything. Angles P and S are interior, so they're supplementary (P + S = 180). Likewise, P and Q are interior. Here you'll notice that angles S and Q are congruent. So you can safely expect that angle R will be congruent to angle P. Check: angle R and angle S are interior (as are R and Q), so if you do the maths, R will come out to 63º as well.
6) Lines PQ and TR are parallel with two possible transversals: PR and QS. If you redraw the diagram without PR, you'll see that angles Q and TRS are corresponding (and what do we know about corresponding angles?). Triangle PQR is isosceles, which means that the angles opposite to the congruent sides will be congruent themselves. So because PQ and PR are congruent line segments, the angles opposite them (angles PRQ and Q respectively) will be congruent as well. So we know two of the angles in the triangle PQR—angles Q and PRQ. The last angle is angle P. If you have already learnt about the sum of angles in a triangle being equal to 180º, use that (angles P + Q + PRQ = 180º, then substitute). Otherwise, use the straight line principle at point R (so that angles PRQ + PRT + TRS = 180º, then substitute). Over transversal PR, the angles P and PRT are alternate interior angles, which are congruent.
I decided to explain the processes I used so that your sister can understand why I'm saying the answer is so. Hope it helps.
