Exam practice

False. Two right angles already make 180∘°, leaving 0∘° for the third angle, so there is no triangle. A triangle has at most one right angle.

False as a rule: it is true only for an acute-angled triangle. A right-angled or obtuse-angled triangle has one angle that is not acute. (What is always true: every triangle has at least two acute angles.)

False. The adjacent interior angle is $180^\circ - 130^\circ = 50^\circ$, which is acute.

True. The exterior angle is 180∘° minus the interior angle; 180∘° minus anything less than 90∘° is more than 90∘°.

y and 80∘° are vertically opposite angles, so $y = 80^\circ$. In the triangle, $x = 180^\circ - 50^\circ - 80^\circ =$ 50∘°.

x is an exterior angle: $x = 50^\circ + 60^\circ = 110^\circ$. On the straight line, $y = 180^\circ - 110^\circ = 70^\circ$.

On the straight line, $y = 180^\circ - 120^\circ = 60^\circ$. The exterior angle 120∘° = 50∘° + x, so $x = 70^\circ$.

y and 90∘° are vertically opposite: $y = 90^\circ$. Then $x + x + 90^\circ = 180^\circ$, so $2x = 90^\circ$ and $x = 45^\circ$.

The exterior angle $(8x - 11)^\circ$ equals the two opposite interior angles:
$$8x - 11 = (2x + 6) + 5x \quad\Rightarrow\quad 8x - 11 = 7x + 6 \quad\Rightarrow\quad x = 17$$
x = 17. Check: the angles are $2 \times 17 + 6 = 40^\circ$ and $5 \times 17 = 85^\circ$, and 40 + 85 = 125 = 8 × 17 − 11.

$$a + 62 = (a - 16) + (a + 22) = 2a + 6 \quad\Rightarrow\quad a = 56$$
a = 56. The angles are 40∘°, 78∘° and 62∘°, and the exterior angle is 118∘° = 40∘° + 78∘°.

$$3x - 10 = 25 + (x + 15) \quad\Rightarrow\quad 2x = 50 \quad\Rightarrow\quad x = 25$$
x = 25. The interior angles at the base are 25∘° and 40∘°, and the exterior angle is 65∘°.

On the straight lines: $y = 180^\circ - 120^\circ = 60^\circ$ and $z = 180^\circ - 130^\circ = 50^\circ$. Then $x = 180^\circ - 60^\circ - 50^\circ = 70^\circ$. (Or: the exterior angle $DCB = 120^\circ = x + z$, so $x = 70^\circ$.)

∠UQP is a right angle, so $2x = 90^\circ$ and $x = 45^\circ$. Then $\angle VPR = 3x = 135^\circ$, so on the straight line VPQ, $\angle QPR = 180^\circ - 135^\circ = 45^\circ$. In triangle PQR, $\angle PQR = 90^\circ$, so $\angle PRQ = 45^\circ$, and $y = \angle PRT = 180^\circ - 45^\circ = 135^\circ$. (Check: $\angle QRS = 3x = 135^\circ$ is vertically opposite y, as it should be.)

In triangle ABC, $\angle ACB = 180^\circ - 45^\circ - 30^\circ = 105^\circ$. ∠DCF is vertically opposite ∠ACB, so $x = 105^\circ$. In triangle CDF, the exterior angle EDF = x + y: $130^\circ = 105^\circ + y$, so $y = 25^\circ$.

Let them be x and 2x: $x + 2x = 120^\circ$, so $x = 40^\circ$. The angles are $40^\circ$, $80^\circ$ and the adjacent one, $180^\circ - 120^\circ = 60^\circ$. 40∘°, 80∘° and 60∘°.

Let the angles be 2k, k and k: $4k = 180^\circ$, so $k = 45^\circ$. Then $\angle BAC = 90^\circ$, $\angle ABC = \angle ACB = 45^\circ$, and $\angle ECD = \angle ABC = 45^\circ$. On the straight line BCD:
$$\angle ACE = 180^\circ - \angle ACB - \angle ECD = 180^\circ - 45^\circ - 45^\circ = 90^\circ$$
$\angle ACE = 90^\circ$. (Equal corresponding angles also tell you $CE \parallel BA$.)

AB = AC, so ∠C = ∠B: $x = 70^\circ$.

AB = AC, so $\angle C = \angle B = 35^\circ$, and $x = 180^\circ - 35^\circ - 35^\circ = 110^\circ$.

MN = NP and $\angle N = 90^\circ$, so $\angle M = \angle P = 45^\circ$. In triangle MNQ, MQ = MN, so $\angle MQN = \angle MNQ = \dfrac{180^\circ - 45^\circ}{2} = 67.5^\circ$. On the straight line MQP: $x = 180^\circ - 67.5^\circ = 112.5^\circ$.

AB = AC and $AD \perp BC$, so AD bisects ∠A: $x = 35^\circ$. In triangle ABD, $y = 180^\circ - 90^\circ - 35^\circ = 55^\circ$, and the base angles are equal: $z = 55^\circ$.

PQ = PR, so $\angle Q = \angle R = 55^\circ$ and $\angle QPR = 180^\circ - 110^\circ = 70^\circ$. $PS \perp QR$ bisects the vertical angle: $x = y = 35^\circ$.

y is vertically opposite 80∘°: $y = 80^\circ$. Then $x = 180^\circ - 50^\circ - 80^\circ = 50^\circ$: the triangle is isosceles, with two 50∘° angles.

At the bottom left, the exterior angle is 140∘°, so the interior angle there is $40^\circ$. The two sides of the lower triangle from the crossing are equal, so its other base angle is also 40∘°, and $x = 180^\circ - 40^\circ - 40^\circ = 100^\circ$. The angle at the crossing in the upper triangle is vertically opposite x, 100∘°; its two sides are equal too, so each of its base angles is $(180^\circ - 100^\circ) \div 2 = 40^\circ$. y is next to one of them on the straight top line: $y = 180^\circ - 40^\circ = 140^\circ$.

In the lower triangle the two sides from the crossing are equal, so its base angles are both 55∘°, and $x = 180^\circ - 110^\circ = 70^\circ$. The angle at the crossing in the upper triangle is vertically opposite, 70∘°, and that triangle is isosceles too: $y = (180^\circ - 70^\circ) \div 2 = 55^\circ$.

$y = 90^\circ$ (vertically opposite), and $2x = 90^\circ$, so $x = 45^\circ$: an isosceles right-angled triangle.

Each angle of the equilateral triangle is 60∘°. At its bottom-right corner the right angle is split into 60∘° and $90^\circ - 60^\circ = 30^\circ$. The second triangle has its two marked sides equal, so its angle at the top corner equals that 30∘°: $x = 30^\circ$, and $y = 180^\circ - 30^\circ - 30^\circ = 120^\circ$.

The left triangle is isosceles with vertical angle 75∘°: $x = (180^\circ - 75^\circ) \div 2 = 52.5^\circ$. So its angle at the crossing is also 52.5∘°, and the right triangle’s angle at the crossing is vertically opposite, 52.5∘°. The right triangle is isosceles, with its equal angles at the crossing and at x (52.5∘° each), so $y = 180^\circ - 105^\circ = 75^\circ$.

The marks say XY = XZ and XP = XQ:
$$3y = 7x \qquad\text{and}\qquad 9x = 13 + 2y$$
From the first, $y = \dfrac{7x}{3}$. Put it in the second: $9x = 13 + \dfrac{14x}{3}$, so 27x = 39 + 14x, 13x = 39 and x = 3; then y = 7. Check: XY = XZ = 21 and XP = XQ = 27.

In triangle ABC, CA = CB, so $\angle CAB = \angle CBA = (180^\circ - 66^\circ) \div 2 = 57^\circ$. In triangle BCD, CB = CD and $\angle BCD = 90^\circ$, so $\angle CBD = \angle CDB = 45^\circ$. On the straight line ABE:
$$\angle DBE = 180^\circ - 57^\circ - 45^\circ = 78^\circ$$
In the right-angled triangle BED: $\angle BDE = 90^\circ - 78^\circ = 12^\circ$.

a. Triangle ACD is isosceles (AC = AD) and the pole AB is perpendicular to its base, so B is the middle of CD: BC = BD. By Pythagoras’ rule in triangle ABC:
$$BC = \sqrt{13^2 - 12^2} = \sqrt{169 - 144} = \sqrt{25} = 5 \text{ m}$$
Each wire is pegged 5 m from the foot of the pole.
b. Yes. $\triangle ABC \cong \triangle ABD$ by R.H.S. (right angles at B, hypotenuses AC = AD, common side AB), so they have the same area: each is $\frac12 \times 5 \times 12 = 30$ m².

Your lengths depend on the size of the drawing; for a copy with BC = 2.8 cm, AB = 3.6 cm and $AC \approx 4.6$ cm. The longest side is AC, so the largest angle is ∠B, the one facing it (a right angle here). The shortest side is BC, so the smallest angle is ∠A (about 38∘°).

The angles measure about $\angle A = 30^\circ$, $\angle B = 129^\circ$ and $\angle C = 21^\circ$ (they add up to 180∘°). The largest angle is at B, so the longest side is AC; the smallest is at C, so the shortest side is AB.

The two shorter sides must add up to more than the longest: 5 + 6 > x, so (b) x < 11. (d) is wrong, because at x = 11 the sides lie flat: no triangle. Since x is the longest side, it is also at least 6: 6 ≤ x < 11.

The height AD from the vertex bisects the base, so BD = 12 cm. From the area:
$$\tfrac12 \times 24 \times AD = 60 \quad\Rightarrow\quad AD = 5 \text{ cm}$$
In the right-angled triangle ABD:
$$AB = \sqrt{12^2 + 5^2} = \sqrt{169} = 13 \text{ cm}$$
Each of the equal sides is 13 cm.

Proved.

Proved. What it means: the straight line PS is shorter than any path from P to S that bends on the way.

By Theorem 1, $\dfrac{AE}{EC} = \dfrac{AD}{DB} = \dfrac23$, so $AE = \dfrac23 \times 12 =$ 8 cm.

By Theorem 1, $\dfrac{XP}{PY} = \dfrac{XQ}{QZ} = \dfrac35$, so $XP = \dfrac35 \times 7.5 = 4.5$ cm, and XY = XP + PY = 4.5 + 7.5 = 12 cm.

$\triangle ABD \sim \triangle ECD$: right angles at B and C, and ∠D is common. BD = 24 + 12 = 36 cm, so
$$\frac{AB}{EC} = \frac{BD}{CD} \quad\Rightarrow\quad \frac{x}{1.8} = \frac{36}{12} = 3$$
x = 5.4 cm.

Each ladder, its wall and the ground make a right-angled triangle, and the angles with the ground are the same, so the triangles are similar:
$$\frac{\text{short ladder}}{60} = \frac{18}{45} \quad\Rightarrow\quad \text{short ladder} = \frac{18 \times 60}{45} = 24$$
The shorter ladder is 24 ft long.

At the same moment, the sun’s rays make the same angle with the ground, so the boy and his shadow, and the tree and its shadow, make similar right-angled triangles:
$$\frac{x}{5} = \frac{14}{7} = 2 \quad\Rightarrow\quad x = 10$$
The tree is 10 ft tall.

$$\frac{h}{20} = \frac{7.5}{30} \quad\Rightarrow\quad h = \frac{7.5 \times 20}{30} = 5$$
The other pole is 5 m high.

This is the opening question. Measure three lengths: the distance from the mirror to the foot of the tree (TM), from the mirror to Mingma’s feet (FM), and the height of his eyes (EF). The tree and Mingma stand upright, and light leaves the mirror at the same angle as it arrives, so the two triangles are similar (Back to the tree):
$$\text{height of the tree} = \frac{EF \times TM}{FM}$$
For example, with TM = 10 m, FM = 2 m and EF = 1.5 m, the tree is $\frac{1.5 \times 10}{2} = 7.5$ m tall. Small errors: the mirror must lie flat, and the eye height is measured to the eyes, not the top of the head.