IB Applications and Interpretation SL · Trigonometry · Section 1 of 4

Prior knowledge: Pythagoras, similarity, arcs and sectors

3 worked examples · By Álvaro López

Key idea

Before using sine, cosine and tangent you need five tools you already know from ESO: the angle sum of a triangle, Pythagoras' theorem, similarity, degrees with minutes and seconds, and the formulas for arc length and sector area. In the IB exam they appear on their own in easy problems, and they are almost always mixed in with other trigonometry parts.

Angles of a triangle

One way to see this: if you tear off the three corners of a paper triangle and place them together at one point, they form a straight line, that is, a half turn.

In a right-angled triangle one of the angles is $9 0^{\circ}$ , so the other two add up to $9 0^{\circ}$ . Two angles that add up to $9 0^{\circ}$ are called complementary: if one measures $α$ , the other measures $9 0^{\circ} - α$ .

Pythagoras' theorem

In a right-angled triangle, the side opposite the right angle is the hypotenuse and is always the longest. The other two sides are the legs.

Right-angled triangle with legs a and b, hypotenuse c and acute angles alpha and beta

To find a leg, rearrange: $a = c^{2} - b^{2}$ . For example, with a hypotenuse of $10$ and a leg of $6$ , the other leg is $100 - 36 = 8$ .

When the sides depend on $x$ . In many problems the sides are given as expressions such as $x$ , $x + 1$ , $x + 9$ . Write Pythagoras' theorem with those expressions, expand the squares and you get a quadratic equation. The hypotenuse is the longest side, so it goes alone on the left-hand side.

Similar triangles

Two triangles are similar when they have the same shape but a different size: one is an enlargement or reduction of the other.

The common value of these ratios is the scale factor: if it is $2$ , every side of the large triangle is twice as long as the corresponding side of the small one. This is how heights that cannot be measured directly are calculated: an object and its shadow form, with the ground and the sun, a triangle similar to that formed by another object and its shadow at the same time of day.

Measuring angles: degrees, minutes and seconds

Angles are measured in sexagesimal degrees: a full turn is $36 0^{\circ}$ . For greater precision, each degree is divided into $60$ minutes and each minute into $60$ seconds: $1^{\circ} = 6 0^{'} 1^{'} = 6 0^{''}$

To convert degrees, minutes and seconds to decimal degrees, divide the minutes by $60$ and the seconds by $3600$ . For example, $2 0^{\circ} 3 0^{'} 1 8^{''} = 20 + \frac{30}{60} + \frac{18}{3600} = 20.50 5^{\circ}$ . To go the other way, multiply the decimal part by $60$ ; this gives the minutes. Multiply the decimal part of the minutes by $60$ to get the seconds.

Arc length and sector area

An arc is a piece of a circumference, and a circular sector is the region between two radii and their arc, like a slice of pizza.

Circular sector with centre O, radius r and angle theta, with arc of length L

The idea is proportional. An angle of $θ^{\circ}$ is the fraction $\frac{θ}{360}$ of a full turn, so the arc is that fraction of the circumference and the sector is that fraction of the circle.

For example, a sector of $9 0^{\circ}$ and radius $4$ is a quarter of the circle: $L = \frac{1}{4} \cdot 2 π \cdot 4 = 2 π$ and $A = \frac{1}{4} \cdot π \cdot 16 = 4 π$ .

Worked examples

Example 1 Basic

Right-angled triangle and angles

a) In a triangle, two angles measure $5 2^{\circ}$ and $7 1^{\circ}$ . Calculate the third.

b) A right-angled triangle has legs of $20 cm$ and $21 cm$ . Calculate the hypotenuse.

c) One of the acute angles of a right-angled triangle measures $3 7^{\circ}$ . Calculate the other.

Solution

a) The three angles add up to $18 0^{\circ}$ : $18 0^{\circ} - 5 2^{\circ} - 7 1^{\circ} = 5 7^{\circ}$ .

b) By Pythagoras, $c^{2} = 2 0^{2} + 2 1^{2} = 400 + 441 = 841$ , so $c = 841 = 29 cm$ .

c) The acute angles are complementary: $9 0^{\circ} - 3 7^{\circ} = 5 3^{\circ}$ .

Example 2 Medium

A shadow and an angle

A person $1.8 m$ tall casts a shadow $2.4 m$ long. At that same moment, a tree casts a shadow $10 m$ long.

a) Calculate the height of the tree.

b) An instrument measures an angle of $3 5^{\circ} 2 4^{'} 3 6^{''}$ . Express it in decimal degrees.

c) Express $72.35 5^{\circ}$ in degrees, minutes and seconds.

Solution

a) At the same time of day the sun's rays arrive at the same angle, so the triangles formed by each object and its shadow are similar (both have a right angle and the same angle on the ground). With $h$ the height of the tree: $\frac{h}{10} = \frac{1.8}{2.4} \Rightarrow h = \frac{1.8 \cdot 10}{2.4} = 7.5 m$

b) $35 + \frac{24}{60} + \frac{36}{3600} = 35 + 0.4 + 0.01 = 35.4 1^{\circ}$ .

c) The whole degrees are $7 2^{\circ}$ . The decimal part $0.35 5^{\circ}$ is multiplied by $60$ : $0.355 \cdot 60 = 21. 3^{'}$ , that is, $2 1^{'}$ and a decimal part $0. 3^{'}$ . Multiply this by $60$ : $0.3 \cdot 60 = 1 8^{''}$ . The result is $7 2^{\circ} 2 1^{'} 1 8^{''}$ .

Example 3 Hard

A park with a triangle and a flower bed

In a park there is a triangular area with a right angle. Its sides measure $x$ , $(x + 1)$ and $(x + 9)$ metres, where $x$ is positive. Next to it a flower bed in the shape of a circular sector is built, with central angle $8 0^{\circ}$ and whose radius is the length, in metres, of the longest side of the triangle.

a) Write a quadratic equation in $x$ and find the value of $x$ .

b) Calculate the arc length of the flower bed, to two decimal places.

c) Calculate the area of the flower bed, to two decimal places.

Solution

a) The longest side is the hypotenuse, $x + 9$ . By Pythagoras: $(x + 9)^{2} = x^{2} + (x + 1)^{2}$ Expanding, $x^{2} + 18 x + 81 = 2 x^{2} + 2 x + 1$ . Moving everything to one side: $x^{2} - 16 x - 80 = 0 \Rightarrow (x - 20) (x + 4) = 0$ The solutions are $x = 20$ and $x = - 4$ . As a length is positive, $x = 20$ . The sides measure $20$ , $21$ and $29$ metres, and $2 0^{2} + 2 1^{2} = 841 = 2 9^{2}$ checks out.

b) The radius is the hypotenuse, $r = 29 m$ . The arc length is $L = \frac{80}{360} \cdot 2 π \cdot 29 = \frac{116 π}{9} \approx 40.49 m$

c) The area of the sector is $A = \frac{80}{360} \cdot π \cdot 2 9^{2} = \frac{2}{9} π \cdot 841 \approx 587.13 m^{2}$

In the exam: marks are given for writing Pythagoras' equation with the hypotenuse in the correct place, solving the quadratic and rejecting the negative solution. For the arc and sector formulas, write the substitution before the result, and round only at the end, without using already-rounded values.

Common mistakes

Summary

The angles of a triangle add up to $18 0^{\circ}$ ; the two acute angles of a right-angled triangle add up to $9 0^{\circ}$ (they are complementary).
Pythagoras: $c^{2} = a^{2} + b^{2}$ , with $c$ the hypotenuse. If the sides depend on $x$ , you get a quadratic: expand correctly and reject negative solutions.
Similar triangles: equal angles and equal ratios between corresponding sides.
$1^{\circ} = 6 0^{'}$ and $1^{'} = 6 0^{''}$ . To convert to decimal: minutes divided by $60$ and seconds divided by $3600$ .
Arc: $L = \frac{θ}{360} \cdot 2 π r$ . Sector area: $A = \frac{θ}{360} \cdot π r^{2}$ . Sector perimeter: $2 r + L$ .
On the GDC, always check that it is in degree mode before using angles.

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