Isosceles and Equilateral Triangles

The base angles of an isosceles triangle are equal, and each angle of an equilateral triangle is 60°. Use these facts, and the same facts the other way round, to find unknown angles and sides in figures.

Before you start

A triangle is isosceles if at least two of its sides are equal, and equilateral if all three of its sides are equal. Equal sides are marked with the same number of small ticks (Triangles and Their Angles).
The angles of a triangle add up to 180°, written (∠ sum of △). If two angles of a triangle are 50° and 60°, the third is $180^{\circ} - 50^{\circ} - 60^{\circ} = 70^{\circ}$ (Triangles and Their Angles).
An exterior angle of a triangle is equal to the sum of the two interior opposite angles, written (ext. ∠ of △) (Triangles and Their Angles).
Adjacent angles on a straight line add up to 180° (adj. ∠s on a str. line). When $AB \parallel CD$ , alternate angles are equal (alt. ∠s, AB // CD) and corresponding angles are equal (corr. ∠s, AB // CD) (Chapter 11).

The Isosceles Triangle

In an isosceles triangle the two equal sides meet at a vertex called the apex. The angle there, between the two equal sides, is the apex angle. The side opposite the apex is the base, and the two angles at the ends of the base are the base angles.

In the figure, $AB = AC$ (one tick on each). $A$ is the apex and $\angle BAC$ is the apex angle; $BC$ is the base, and $\angle ABC$ and $\angle ACB$ are the base angles.

The base is fixed by the equal sides, not by the way the triangle is drawn: it is always the side opposite the apex, even when the triangle is drawn standing on one of its equal sides.

Base angles of an isosceles triangle

The base angles of an isosceles triangle are equal.

In the figure: if $AB = AC$ , then $\angle ABC = \angle ACB$ .

The reason is written (base ∠s of isos. △), short for "base angles of an isosceles triangle".

Why the base angles are equal

Cut out an isosceles triangle $ABC$ with $AB = AC$ . Fold it through $A$ so that side $AB$ lies along side $AC$ .

Because $AB = AC$ , the point $B$ lands exactly on $C$ , and the fold line meets $BC$ at its midpoint. The two halves of the triangle now lie exactly on each other.

So $\angle ABC$ lies exactly on $\angle ACB$ : the base angles are equal. This fold line is a line of symmetry of the triangle. Lines of symmetry, and turning a figure so that it fits onto itself, are taught in the next lesson, Line and Rotational Symmetry.

Because the base angles are equal, one angle of an isosceles triangle gives you the other two.

Given the apex angle, take it from 180° and halve what is left: each base angle $= (180^{\circ} - \text{apex angle}) \div 2$ .

Given a base angle, the other base angle is the same, so take both from 180°: apex angle $= 180^{\circ} - 2 \times \text{base angle}$ .

Example 1: The base angles from the apex angle

In $\triangle ABC$ , $AB = AC$ and $\angle BAC = 48^{\circ}$ . Find $\angle ABC$ and $\angle ACB$ .

Solution

$AB = AC$ , so $\triangle ABC$ is isosceles with apex $A$ , and $\angle ABC = \angle ACB$ (base ∠s of isos. △).

The two base angles together make $180^{\circ} - 48^{\circ} = 132^{\circ}$ (∠ sum of △).

So each base angle is $132^{\circ} \div 2 = 66^{\circ}$ .

Check: $48^{\circ} + 66^{\circ} + 66^{\circ} = 180^{\circ}$ .

$\angle ABC = \angle ACB = 66^{\circ}$ .

Example 2: The apex angle from a base angle

In $\triangle ABC$ , $AB = AC$ and $\angle ABC = 53^{\circ}$ . Find $x$ .

Solution

$AB = AC$ , so the base angles are at $B$ and $C$ :

$\angle ACB = \angle ABC = 53^{\circ}$ (base ∠s of isos. △)

$x = 180 - 53 - 53 = 74$ (∠ sum of △)

Check: $74^{\circ} + 53^{\circ} + 53^{\circ} = 180^{\circ}$ .

$x = 74$ , so the apex angle $\angle BAC = 74^{\circ}$ .

The rule also works the other way round. If two angles of a triangle are equal, the triangle folds onto itself in the same way, and the two sides opposite the equal angles are equal.

The side opposite an angle is the side that does not touch that angle's vertex. In $\triangle PQR$ , the side opposite $\angle PQR$ is $PR$ , and the side opposite $\angle PRQ$ is $PQ$ .

Equal angles give equal sides

If two angles of a triangle are equal, then the sides opposite them are equal, and the triangle is isosceles.

In $\triangle PQR$ : if $\angle PQR = \angle PRQ$ , then $PR = PQ$ .

The reason is written (isos. △).

Example 3: Finding the equal sides from the angles

In $\triangle XYZ$ , $\angle ZXY = 40^{\circ}$ , $\angle XYZ = 70^{\circ}$ and $XY = 9$ cm.

(a) Find $\angle XZY$ .

(b) Which two sides of the triangle are equal? Give the length of the side that is equal to $XY$ .

Solution

(a) $\angle XZY = 180^{\circ} - 40^{\circ} - 70^{\circ} = 70^{\circ}$ (∠ sum of △)

(b) $\angle XYZ = \angle XZY = 70^{\circ}$ , so the sides opposite these two angles are equal (isos. △).

The side opposite $\angle XYZ$ (the angle at $Y$ ) is $XZ$ . The side opposite $\angle XZY$ (the angle at $Z$ ) is $XY$ .

So $XZ = XY = 9$ cm.

The equal sides $XY$ and $XZ$ meet at $X$ , so $X$ is the apex and $YZ$ is the base, although the triangle is drawn standing on $XY$ . The side $YZ$ , which joins the two equal angles, is not one of the equal sides.

(a) $\angle XZY = 70^{\circ}$ (b) $XY = XZ$ , so $XZ = 9$ cm.

The Equilateral Triangle

An equilateral triangle has all three sides equal, so it is isosceles in three ways: whichever vertex you take as the apex, the two sides that meet there are equal.

So each pair of its angles is a pair of base angles of an isosceles triangle, and all three angles are equal (base ∠s of isos. △). Three equal angles that add up to 180° are each $180^{\circ} \div 3 = 60^{\circ}$ .

Angles of an equilateral triangle

Each angle of an equilateral triangle is 60°.

Conversely, if all three angles of a triangle are equal (so each is 60°), then all three sides are equal, and the triangle is equilateral.

The reason, either way, is written (equilateral △).

Why three equal angles give three equal sides

In $\triangle ABC$ , let $\angle BAC = \angle ABC = \angle ACB$ .

Since $\angle BAC = \angle ABC$ , the sides opposite them are equal: $BC = AC$ (isos. △).

Since $\angle ABC = \angle ACB$ , the sides opposite them are equal: $AC = AB$ (isos. △).

So $AB = BC = CA$ , and the triangle is equilateral.

One useful result follows: an isosceles triangle with one angle of 60° is equilateral. If its apex angle is 60°, each base angle is $(180^{\circ} - 60^{\circ}) \div 2 = 60^{\circ}$ . If a base angle is 60°, the apex angle is $180^{\circ} - 2 \times 60^{\circ} = 60^{\circ}$ . Either way, all three angles are 60°.

Example 4: An isosceles triangle with a 60° angle

In $\triangle ABC$ , $AB = AC$ , $\angle ABC = 60^{\circ}$ and $BC = 5$ cm.

(a) Find $\angle ACB$ and $\angle BAC$ .

(b) Find the length of $AB$ .

Solution

(a) $\angle ACB = \angle ABC = 60^{\circ}$ (base ∠s of isos. △)

$\angle BAC = 180^{\circ} - 60^{\circ} - 60^{\circ} = 60^{\circ}$ (∠ sum of △)

(b) All three angles are 60°, so $\triangle ABC$ is equilateral (equilateral △), and $AB = BC = 5$ cm.

Check: $60^{\circ} + 60^{\circ} + 60^{\circ} = 180^{\circ}$ .

(a) $\angle ACB = 60^{\circ}$ and $\angle BAC = 60^{\circ}$ (b) $AB = 5$ cm.

Finding Unknown Angles

In a figure with several lines, look for the ticks first: every pair of equal sides gives a pair of equal base angles, at the two ends of the third side (the base). Write the reason after every step, and check that the angles of each triangle add up to 180°.

Example 5: With an exterior angle

In the figure, $AB = AC$ and $BC$ is produced to $D$ . $\angle ACD = 115^{\circ}$ . Find $x$ .

Solution

$\angle ACB = 180^{\circ} - 115^{\circ} = 65^{\circ}$ (adj. ∠s on a str. line)

$\angle ABC = \angle ACB = 65^{\circ}$ (base ∠s of isos. △)

$x = 180 - 65 - 65 = 50$ (∠ sum of △)

Another way: $\angle ACD$ is an exterior angle of $\triangle ABC$ , so $115 = x + 65$ (ext. ∠ of △), which gives $x = 50$ again.

Check: $50^{\circ} + 65^{\circ} + 65^{\circ} = 180^{\circ}$ , and $65^{\circ} + 115^{\circ} = 180^{\circ}$ on the line $BD$ .

$x = 50$ , so $\angle BAC = 50^{\circ}$ .

Example 6: Two isosceles triangles sharing a side

In the figure, $B$ , $C$ and $D$ lie on a straight line, $AB = AC = CD$ and $\angle ABC = 70^{\circ}$ .

Find (a) $\angle BAC$ , (b) $\angle CAD$ and (c) $\angle BAD$ .

Solution

(a) In $\triangle ABC$ , $AB = AC$ , so $\angle ACB = \angle ABC = 70^{\circ}$ (base ∠s of isos. △).

$\angle BAC = 180^{\circ} - 70^{\circ} - 70^{\circ} = 40^{\circ}$ (∠ sum of △)

(b) $\angle ACD = 180^{\circ} - 70^{\circ} = 110^{\circ}$ (adj. ∠s on a str. line)

In $\triangle ACD$ , $CA = CD$ , so $C$ is its apex and its base angles are $\angle CAD$ and $\angle CDA$ : $\angle CAD = \angle CDA$ (base ∠s of isos. △).

$\angle CAD = (180^{\circ} - 110^{\circ}) \div 2 = 35^{\circ}$ (∠ sum of △)

Check in the large triangle $ABD$ : $\angle ABD + \angle BAD + \angle ADB = 70^{\circ} + 75^{\circ} + 35^{\circ} = 180^{\circ}$ .

(a) $\angle BAC = 40^{\circ}$ (b) $\angle CAD = 35^{\circ}$ (c) $\angle BAD = 75^{\circ}$ .

Example 7: With parallel lines

In the figure, $XAY$ is a straight line parallel to $BC$ , $AB = AC$ and $\angle XAB = 68^{\circ}$ . Find $x$ .

Solution

$\angle ABC = \angle XAB = 68^{\circ}$ (alt. ∠s, XY // BC)

$\angle ACB = \angle ABC = 68^{\circ}$ (base ∠s of isos. △)

$x = 180 - 68 - 68 = 44$ (∠ sum of △)

Check: $\angle YAC = \angle ACB = 68^{\circ}$ (alt. ∠s, XY // BC), and the three angles at $A$ on the line $XY$ add up to $68^{\circ} + 44^{\circ} + 68^{\circ} = 180^{\circ}$ .

$x = 44$ , so $\angle BAC = 44^{\circ}$ .

Example 8: Base angles in terms of x

In $\triangle ABC$ , $AB = AC$ , $\angle ABC = (3x - 10)^{\circ}$ and $\angle ACB = (2x + 15)^{\circ}$ . Find $x$ and $\angle BAC$ .

Solution

$AB = AC$ , so the base angles are equal (base ∠s of isos. △):

$3x - 10 = 2x + 15$

Subtract $2x$ from both sides: $x - 10 = 15$

Add 10 to both sides: $x = 25$

So $\angle ABC = (3 \times 25 - 10)^{\circ} = 65^{\circ}$ and $\angle ACB = (2 \times 25 + 15)^{\circ} = 65^{\circ}$ : equal, as they should be.

$\angle BAC = 180^{\circ} - 65^{\circ} - 65^{\circ} = 50^{\circ}$ (∠ sum of △)

Check: $65^{\circ} + 65^{\circ} + 50^{\circ} = 180^{\circ}$ .

$x = 25$ and $\angle BAC = 50^{\circ}$ .

Common Mistakes

Taking the wrong angles as equal. When $AB = AC$ , the equal angles are at $B$ and $C$ , the ends of the base, not at the apex $A$ . Find the base by the ticks, not by which side is drawn at the bottom.
Forgetting to halve, or to take away both base angles. If the apex angle is 40°, each base angle is $(180^{\circ} - 40^{\circ}) \div 2 = 70^{\circ}$ , not 140°. If a base angle is 40°, the apex angle is $180^{\circ} - 2 \times 40^{\circ} = 100^{\circ}$ , not 140°.
Naming the wrong equal sides. When two angles are equal, the equal sides are the sides opposite them. In $\triangle XYZ$ with $\angle XYZ = \angle XZY$ , the equal sides are $XZ$ and $XY$ ; the side $YZ$ between the two equal angles is the base.
Deciding that a triangle is isosceles because it looks it. Use only what is marked or given: ticks, equal lengths or equal angles.

Practice

Try each question, then open its solution to check your working. The figures are drawn to scale.

Practice 1: Base angles

In $\triangle PQR$ , $PQ = PR$ and $\angle QPR = 34^{\circ}$ . Find $x$ .

ShowHide solution

$PQ = PR$ , so $\angle PQR = \angle PRQ = x^{\circ}$ (base ∠s of isos. △).

$x + x = 180 - 34 = 146$ (∠ sum of △)

$x = 146 \div 2 = 73$

Check: $34^{\circ} + 73^{\circ} + 73^{\circ} = 180^{\circ}$ .

$x = 73$ , so $\angle PQR = 73^{\circ}$ .

Practice 2: An obtuse apex angle

In $\triangle KLM$ , $KL = KM$ and $\angle KLM = 28^{\circ}$ . Find $x$ .

ShowHide solution

$KL = KM$ , so the base angles are at $L$ and $M$ : $\angle KML = \angle KLM = 28^{\circ}$ (base ∠s of isos. △).

$x = 180 - 28 - 28 = 124$ (∠ sum of △)

Check: $124^{\circ} + 28^{\circ} + 28^{\circ} = 180^{\circ}$ .

$x = 124$ , so $\angle LKM = 124^{\circ}$ , an obtuse angle.

Practice 3: A pair of compasses

Wei Ming opens his compasses and stands them on a sheet of paper. The two arms $AB$ and $AC$ are equal, the hinge is at $A$ , and the point $B$ and the pencil tip $C$ rest on the paper. The angle between the arms is 44°.

(a) Find $x$ , the angle each arm makes with the paper.

(b) Wei Ming now opens the compasses wider, until $B$ and $C$ are as far apart as the length of each arm. What is the angle between the arms now?

ShowHide solution

(a) $AB = AC$ , so $\angle ACB = \angle ABC = x^{\circ}$ (base ∠s of isos. △).

$x + x = 180 - 44 = 136$ (∠ sum of △)

$x = 136 \div 2 = 68$

Check: $44^{\circ} + 68^{\circ} + 68^{\circ} = 180^{\circ}$ .

(b) Now $AB = AC = BC$ : all three sides of $\triangle ABC$ are equal, so it is equilateral, and each of its angles is 60° (equilateral △).

(a) $x = 68$ : each arm makes an angle of 68° with the paper. (b) The angle between the arms is 60°.

Practice 4: Which sides are equal?

In $\triangle DEF$ , $\angle EDF = 50^{\circ}$ , $\angle DEF = 80^{\circ}$ and $DE = 6.4$ cm.

(a) Find $\angle DFE$ .

(b) Name the two equal sides of the triangle, and give the length of the one that is not $DE$ .

ShowHide solution

(a) $\angle DFE = 180^{\circ} - 50^{\circ} - 80^{\circ} = 50^{\circ}$ (∠ sum of △)

(b) $\angle EDF = \angle DFE = 50^{\circ}$ , so the sides opposite them are equal (isos. △).

The side opposite $\angle EDF$ (at $D$ ) is $EF$ ; the side opposite $\angle DFE$ (at $F$ ) is $DE$ . So $EF = DE = 6.4$ cm.

The third side, $DF$ , is opposite the 80° angle and is not equal to the other two.

(a) $\angle DFE = 50^{\circ}$ (b) $DE = EF$ , so $EF = 6.4$ cm.

Practice 5: Base angles with x

In $\triangle ABC$ , $AB = AC$ , $\angle ABC = (2x + 9)^{\circ}$ and $\angle ACB = (4x - 23)^{\circ}$ . Find $x$ and $\angle BAC$ .

ShowHide solution

$AB = AC$ , so the base angles are equal (base ∠s of isos. △):

$4x - 23 = 2x + 9$

Subtract $2x$ from both sides: $2x - 23 = 9$

Add 23 to both sides: $2x = 32$

Divide both sides by 2: $x = 16$

So $\angle ABC = (2 \times 16 + 9)^{\circ} = 41^{\circ}$ and $\angle ACB = (4 \times 16 - 23)^{\circ} = 41^{\circ}$ .

$\angle BAC = 180^{\circ} - 41^{\circ} - 41^{\circ} = 98^{\circ}$ (∠ sum of △)

Check: $41^{\circ} + 41^{\circ} + 98^{\circ} = 180^{\circ}$ .

$x = 16$ and $\angle BAC = 98^{\circ}$ .

Practice 6: Equal sides in terms of x

In $\triangle PQR$ , $\angle PQR = \angle PRQ$ , $PQ = (2x + 5)$ cm, $PR = (4x - 3)$ cm and $QR = 10$ cm. Find $x$ and the length of $PQ$ .

ShowHide solution

$\angle PQR = \angle PRQ$ , so the sides opposite them are equal (isos. △). The side opposite $\angle PQR$ is $PR$ , and the side opposite $\angle PRQ$ is $PQ$ . So $PR = PQ$ :

$4x - 3 = 2x + 5$

Subtract $2x$ from both sides: $2x - 3 = 5$

Add 3 to both sides: $2x = 8$

Divide both sides by 2: $x = 4$

$PQ = 2 \times 4 + 5 = 13$ cm. Check: $PR = 4 \times 4 - 3 = 13$ cm, the same.

$QR = 10$ cm joins the two equal angles: it is the base, and it does not enter the equation.

$x = 4$ and $PQ = 13$ cm.

Practice 7: Three angles in x

The angles of $\triangle ABC$ are $\angle BAC = (x + 20)^{\circ}$ , $\angle ABC = (2x - 20)^{\circ}$ and $\angle ACB = (3x - 60)^{\circ}$ , and $AB = 8$ cm.

(a) Find $x$ .

(b) Show that $\triangle ABC$ is equilateral, and find the length of $BC$ .

ShowHide solution

(a) $(x + 20) + (2x - 20) + (3x - 60) = 180$ (∠ sum of △)

Collect like terms: $6x - 60 = 180$

Add 60 to both sides: $6x = 240$

Divide both sides by 6: $x = 40$

(b) $\angle BAC = (40 + 20)^{\circ} = 60^{\circ}$ , $\angle ABC = (2 \times 40 - 20)^{\circ} = 60^{\circ}$ and $\angle ACB = (3 \times 40 - 60)^{\circ} = 60^{\circ}$ .

All three angles are equal, so the triangle is equilateral (equilateral △), and $BC = AB = 8$ cm.

Check: $60^{\circ} + 60^{\circ} + 60^{\circ} = 180^{\circ}$ .

(a) $x = 40$ (b) each angle is 60°, so $\triangle ABC$ is equilateral and $BC = 8$ cm.

Practice 8: An exterior angle at the apex

In the figure, $AB = AC$ and $BA$ is produced to $E$ . $\angle CAE = 110^{\circ}$ . Find $x$ .

ShowHide solution

$AB = AC$ , so $\angle ACB = \angle ABC = x^{\circ}$ (base ∠s of isos. △).

$\angle CAE$ is an exterior angle of $\triangle ABC$ , so it equals the sum of the two interior opposite angles:

$x + x = 110$ (ext. ∠ of △)

$2x = 110$ , so $x = 55$ .

Check: $\angle BAC = 180^{\circ} - 110^{\circ} = 70^{\circ}$ (adj. ∠s on a str. line), and $55^{\circ} + 55^{\circ} + 70^{\circ} = 180^{\circ}$ .

$x = 55$ , so $\angle ABC = 55^{\circ}$ .

Practice 9: A line parallel to the base

In the figure, $AB = AC$ . $D$ lies on $AB$ and $E$ lies on $AC$ , with $DE \parallel BC$ and $\angle ADE = 64^{\circ}$ .

(a) Find $\angle ABC$ and $\angle BAC$ .

(b) Show that $\triangle ADE$ is isosceles, and name its two equal sides.

ShowHide solution

(a) $\angle ABC = \angle ADE = 64^{\circ}$ (corr. ∠s, DE // BC)

$\angle ACB = \angle ABC = 64^{\circ}$ (base ∠s of isos. △)

$\angle BAC = 180^{\circ} - 64^{\circ} - 64^{\circ} = 52^{\circ}$ (∠ sum of △)

(b) $\angle AED = \angle ACB = 64^{\circ}$ (corr. ∠s, DE // BC)

So in $\triangle ADE$ , $\angle ADE = \angle AED = 64^{\circ}$ , and the sides opposite them are equal (isos. △): the side opposite $\angle ADE$ is $AE$ and the side opposite $\angle AED$ is $AD$ . So $AD = AE$ .

Check in $\triangle ADE$ : $64^{\circ} + 64^{\circ} + 52^{\circ} = 180^{\circ}$ .

(a) $\angle ABC = 64^{\circ}$ and $\angle BAC = 52^{\circ}$ (b) $\angle ADE = \angle AED = 64^{\circ}$ , so $\triangle ADE$ is isosceles with $AD = AE$ .

Practice 10: The side of a footbridge

The side of a footbridge over a canal is a frame of steel bars. The top bar $XY$ is parallel to the bottom bar $BD$ . $A$ and $E$ lie on $XY$ , $C$ lies on $BD$ , and $AB = AC = CE$ . $\angle ABC = 58^{\circ}$ .

Find (a) $\angle BAC$ , (b) $\angle CAE$ and (c) $\angle ACE$ .

ShowHide solution

(a) In $\triangle ABC$ , $AB = AC$ , so $\angle ACB = \angle ABC = 58^{\circ}$ (base ∠s of isos. △).

$\angle BAC = 180^{\circ} - 58^{\circ} - 58^{\circ} = 64^{\circ}$ (∠ sum of △)

(b) $\angle CAE = \angle ACB = 58^{\circ}$ (alt. ∠s, XY // BD)

(c) In $\triangle ACE$ , $CA = CE$ , so $C$ is its apex and its base angles are at $A$ and $E$ : $\angle CEA = \angle CAE = 58^{\circ}$ (base ∠s of isos. △).

$\angle ACE = 180^{\circ} - 58^{\circ} - 58^{\circ} = 64^{\circ}$ (∠ sum of △)

Check at $C$ on the line $BD$ : $\angle ECD = \angle AEC = 58^{\circ}$ (alt. ∠s, XY // BD), and $\angle ACB + \angle ACE + \angle ECD = 58^{\circ} + 64^{\circ} + 58^{\circ} = 180^{\circ}$ .

(a) $\angle BAC = 64^{\circ}$ (b) $\angle CAE = 58^{\circ}$ (c) $\angle ACE = 64^{\circ}$ .

Practice 11: Three equal lengths

In the figure, $D$ lies on $AC$ , $AB = BD = DC$ and $\angle ACB = 31^{\circ}$ .

Find (a) $\angle DBC$ , (b) $\angle ADB$ , (c) $\angle BAD$ and (d) $\angle ABC$ .

ShowHide solution

(a) In $\triangle BDC$ , $DB = DC$ , so $\angle DBC = \angle DCB = 31^{\circ}$ (base ∠s of isos. △).

(b) $\angle ADB$ is an exterior angle of $\triangle BDC$ :

$\angle ADB = \angle DBC + \angle DCB = 31^{\circ} + 31^{\circ} = 62^{\circ}$ (ext. ∠ of △)

(d) $\angle ABD = 180^{\circ} - 62^{\circ} - 62^{\circ} = 56^{\circ}$ (∠ sum of △)

$\angle ABC = \angle ABD + \angle DBC = 56^{\circ} + 31^{\circ} = 87^{\circ}$

Check in the large triangle $ABC$ : $\angle BAC + \angle ABC + \angle ACB = 62^{\circ} + 87^{\circ} + 31^{\circ} = 180^{\circ}$ .

(a) $\angle DBC = 31^{\circ}$ (b) $\angle ADB = 62^{\circ}$ (c) $\angle BAD = 62^{\circ}$ (d) $\angle ABC = 87^{\circ}$ .

Practice 12: A roof frame with a strut

The roof frame of a park shelter is a triangle $ABC$ . The two sloping beams $AB$ and $AC$ are equal, the beam $BC$ is horizontal, and each sloping beam makes an angle of 30° with $BC$ . A strut $AD$ joins the top $A$ to the point $D$ on $BC$ , and $AD = DC$ .

Find (a) $\angle BAC$ , (b) $\angle DAC$ and (c) $\angle ADB$ . (d) Show that the strut $AD$ is at right angles to the beam $AB$ .

ShowHide solution

(a) $AB = AC$ , so $\angle ACB = \angle ABC = 30^{\circ}$ (base ∠s of isos. △).

$\angle BAC = 180^{\circ} - 30^{\circ} - 30^{\circ} = 120^{\circ}$ (∠ sum of △)

(b) In $\triangle ADC$ , $DA = DC$ , so $D$ is its apex and its base angles are at $A$ and $C$ : $\angle DAC = \angle DCA = 30^{\circ}$ (base ∠s of isos. △).

$\angle ADB = \angle DAC + \angle DCA = 30^{\circ} + 30^{\circ} = 60^{\circ}$ (ext. ∠ of △)

(d) $\angle BAD = \angle BAC - \angle DAC = 120^{\circ} - 30^{\circ} = 90^{\circ}$ , so $AD$ is at right angles to $AB$ .

Check in $\triangle ABD$ : $\angle ABD + \angle BAD + \angle ADB = 30^{\circ} + 90^{\circ} + 60^{\circ} = 180^{\circ}$ .

(a) $\angle BAC = 120^{\circ}$ (b) $\angle DAC = 30^{\circ}$ (c) $\angle ADB = 60^{\circ}$ (d) $\angle BAD = 90^{\circ}$ , so the strut is at right angles to $AB$ .

Key Points

In an isosceles triangle the equal sides meet at the apex; the side opposite the apex is the base.
The base angles of an isosceles triangle are equal (base ∠s of isos. △).
If two angles of a triangle are equal, the sides opposite them are equal (isos. △).
Each base angle $= (180^{\circ} - \text{apex angle}) \div 2$ ; apex angle $= 180^{\circ} - 2 \times \text{base angle}$ .
Each angle of an equilateral triangle is 60°; a triangle with three equal angles is equilateral (equilateral △).
An isosceles triangle with one angle of 60° is equilateral.
In a figure, find the equal sides from the ticks, give a reason for every step, and check that each triangle's angles add up to 180°.