How to Build a 75 Degree Angle Out of Two Others

By bisecting the gap between them. Construct $60°$ and $90°$ from the same point, then bisect the $30°$ space left between the two arms:

$$\frac{60+90}{2}=75°$$

Every "awkward" angle in the syllabus is built that way. This page covers everything in the ICSE Class 7 Mathematics chapter's second part: the angle pairs made by a transversal, parallel-line calculations, constructing standard angles, and bisectors and parallels.

A transversal is a line cutting two other lines. It creates eight angles, grouped into four named pairs.

- Corresponding angles — in matching positions at the two crossings, one above each line on the same side. They form an F shape.
- Alternate interior angles — between the two lines, on opposite sides of the transversal. They form a Z shape.
- Alternate exterior angles — outside the two lines, on opposite sides of the transversal.
- Co-interior (allied) angles — between the two lines, on the same side of the transversal. They form a U or C shape.

The letter shapes are a reliable way to spot them in a cluttered diagram: look for the F, the Z and the U.

Railway tracks crossed by a sleeper, or the rungs of a ladder against its two rails, are transversals in ordinary life.

These names describe positions only, and that is the point students miss. The pairs exist whether or not the two lines are parallel — what changes with parallelism is whether the angles are equal. So "corresponding angles" is a statement about where an angle sits, not about its size.

When the two lines are parallel, three relationships hold:

- Corresponding angles are equal
- Alternate interior angles are equal (and alternate exterior too)
- Co-interior angles are supplementary, adding to $180°$

Worked example. A transversal cuts two parallel lines and one angle is $70°$. Then the corresponding angle is $70°$, the alternate interior angle is $70°$, and the co-interior angle is

$$180-70=110°$$

All eight angles are therefore either $70°$ or $110°$.

Worked example with algebra. Two co-interior angles are $x$ and $2x$:

$$x+2x=180\Rightarrow 3x=180\Rightarrow x=60°$$

so the angles are $60°$ and $120°$.

Another. Alternate interior angles are $(3x-10)°$ and $(2x+20)°$. Being equal:

$$3x-10=2x+20\Rightarrow x=30$$

so each is $80°$.

Deciding whether two lines are parallel reverses the same rules. If a pair of corresponding angles is equal, or a pair of alternate angles is equal, or a pair of co-interior angles adds to $180°$, then the lines are parallel. If none of these holds, they are not.

So with co-interior angles of $100°$ and $90°$, the lines cannot be parallel, since $100+90=190$ rather than $180$.

Two angles are built from scratch, and the rest come from bisecting.

**$60°$. Draw a base line and mark point O. With any radius, draw an arc from O cutting the line at P. With the same radius** and the point at P, draw an arc crossing the first at Q. Then $\angle QOP=60°$, because OPQ is an equilateral triangle.

**$90°$.** Construct a perpendicular at O. Equivalently, extend the $60°$ method: mark a second arc step to reach $120°$, then bisect between $60°$ and $120°$.

Everything else is halving:

$$30°=\frac{60}{2},45°=\frac{90}{2},15°=\frac{30}{2}$$

$$75°=\frac{60+90}{2},105°=\frac{90+120}{2}$$

So for **$75°$**: construct both $60°$ and $90°$ at O, then bisect the angle between those two arms. For **$105°$**: construct $90°$ and $120°$, then bisect the angle between them.

Folding a square sheet of paper corner to corner gives $45°$ by exactly the same halving idea.

The limit worth knowing is which angles this can reach. Repeated bisection of $60°$ and $90°$ gives $15°$, $30°$, $45°$, $75°$, $105°$ and so on — all multiples of $15°$. An angle such as $20°$ or $50°$ cannot be constructed with ruler and compasses alone, and needs a protractor.

Three constructions complete the chapter, and each uses arcs of equal radius.

Angle bisector. With the point at the vertex B, draw an arc cutting both arms at P and Q. Then from P and Q in turn, with the same radius, draw two arcs crossing at R. The ray BR bisects $\angle B$. So bisecting a $70°$ angle gives two angles of $35°$.

Perpendicular bisector of a segment. Open the compasses to more than half of AB. From A, draw arcs above and below the line; from B with the same opening, repeat. Join the two crossing points. The line cuts AB at its midpoint at $90°$, so an 8 cm segment is split into two of 4 cm.

Parallel line through an external point. Draw any transversal from the point P down to the given line, meeting it at Q. At Q, draw an arc across both the line and the transversal. With the same radius, draw a matching arc at P. Transfer the width of the arc's cut at Q up to the arc at P and mark the point. Join P to it — the line is parallel, because you have copied a corresponding angle.

That last construction depends directly on the earlier section. Equal corresponding angles guarantee parallel lines, so copying the angle is what makes the new line parallel.

And for the perpendicular bisector, an opening of less than half AB makes the arcs fall short of each other, so nothing crosses — arcs that refuse to meet mean the compasses are too tight, not that the drawing is wrong.