Trigonometric Ratios

1. ΔABF = ΔAEC by SAS. This is because AE = AB (because ABDE is a square), AF = AC (because AFGC is a square), and <BAF = <BAC + <CAF = <CAB + <BAE = <CAE. <CAE = <BAF. Let's say triangle ABF has base AF and (if we extend line FA) the altitude from B is equal to AC (they are parallel). Its area then equals half that of square GFAC. On the other triangle, triangle AEC has AE (extended) base and the altitude from C equal to AM (C and M are both on line CL), where M is the point of intersection of AB with the line CL parallel to AE. Thus, the area of ΔAEC equals half that of the rectangle AELM (The base and height are the same as AELM's, so then 1/2 AELM = CAE). That says that the area AC*FA of the square on side AC equals the area of the rectangle AELM. Similarly, the area BC*KB of the square on side BC equals that of rectangle BMLD. Finally, the two rectangles AELM and BMLD make up the square on the original right triangle's hypotenuse AB. So when you have 2 halves of 2 rectangles that make up a square, you get a half of a square. Then, since we started with 1/2 of the area of the triangles, 1/2a^2 + 1/2b^2 = 1/2c^2. Divide by 1/2 and you have the Pythagorean Theorem, a^2 = b^2 = c^2.

2. History

2.1. Pythagoras (570-495 BC): Ionian and Greek philosopher

2.1.1. Although the theorem is known as Pythagoras', when authors Plutarch and Cicero attributed the theorem to Pythagoras, they did so in a way that suggests that the attribution was widely known and undoubted.

2.1.2. Pythagoras' theorem: He studied right-angled triangles and found that the two shorter sides of the triangle squared and then added together equal the square of the longer side (hypotenuse).

3. Water Proof

3.1. The Bride's Chair (Euclid's Proof)

3.2. Start with two squares with sides a and b, respectively, placed side by side. The total area of the two squares is a²+b². The construction did not start with a triangle but now we draw two of them, both with sides a and b and hypotenuse c. Note that the segment common to the two squares has been removed. At this point we therefore have two triangles and a strange looking shape. As a last step, we rotate the triangles 90°, each around its top vertex. The right one is rotated clockwise whereas the left triangle is rotated counterclockwise. Obviously the resulting shape is a square with the side c and area c². This proof appears in a dynamic incarnation.

3.3. Now start with four copies of the same triangle. Three of these have been rotated 90°, 180°, and 270°, respectively. Each has area ab/2. Put them together without additional rotations so that they form a square with side c. The square has a square hole with the side (a - b). Summing up its area (a - b)² and 2ab, the area of the four triangles (4·ab/2), we get C² = (a - b)² + 2ab = a² - 2ab + b² + 2ab = a² + b²

3.4. ΔABF = ΔAEC by SAS. This is because AE = AB (because ABDE is a square), AF = AC (because AFGC is a square), and <BAF = <BAC + <CAF = <CAB + <BAE = <CAE. <CAE = <BAF. Let's say triangle ABF has base AF and (if we extend line FA) the altitude from B is equal to AC (they are parallel). Its area then equals half that of square GFAC. On the other triangle, triangle AEC has AE (extended) base and the altitude from C equal to AM (C and M are both on line CL), where M is the point of intersection of AB with the line CL parallel to AE. Thus, the area of ΔAEC equals half that of the rectangle AELM (The base and height are the same as AELM's, so then 1/2 AELM = CAE). That says that the area AC*FA of the square on side AC equals the area of the rectangle AELM. Similarly, the area BC*KB of the square on side BC equals that of rectangle BMLD. Finally, the two rectangles AELM and BMLD make up the square on the original right triangle's hypotenuse AB. So when you have 2 halves of 2 rectangles that make up a square, you get a half of a square. Then, since we started with 1/2 of the area of the triangles, 1/2a^2 + 1/2b^2 = 1/2c^2. Divide by 1/2 and you have the Pythagorean Theorem, a^2 = b^2 = c^2.

4. TOA CAH SOH

4.1. SOH stands for Sine equals Opposite over Hypotenuse.

4.2. CAH stands for Cosine equals Adjacent over Hypotenuse.

4.3. TOA stands for Tangent equals Opposite over Adjacent.

4.4. Example: Find the values of sin θ, comins θ, and tan θ in the right triangle shown.

4.5. Answer: sin θ = 3/5 = 0.6 cosθ = 4/5 = 0.8 tanθ = 3/4 = 0.75

5. What is it?

5.1. a²+b²=c²

5.2. a and b are known as the two "legs", and are the sides which make up the right angle, while c is known as the hypotenuse, the line opposite from the right angle.

5.3. The Pythagoras' theorem is only true in right-angled triangles, although corollaries of the theorem can be used to determine whether a triangle is acute, obtuse of right-angled.

6. Converse

6.1. Since the pythagoras theorem is only true in the case that the triangle is right-angled, the the presence of a2+b2=c2 in a triangle can be used to determine whether or not the triangle is right angled.

6.2. A corollary of the Pythagorean theorem's converse is allows one to utilise the converse to determine whether a triangle is right, obtuse, or acute, as follows. Let c be chosen to be the longest of the three sides and a + b > c. The following statements apply:

6.2.1. If a2+b2=c2, then the triangle is right.

6.2.2. If a2+b2>c2, then the triangle is acute.

6.2.3. If a2+b2<c2, then the triangle is obtuse.

7. Applications

7.1. Clinometer

7.1.1. A clinometer is used to measure an object's angle of elevation from the ground in a right-angled triangle. This value, when paired with the distance from the object, can then be used to calculate the height of elevation. This is especially useful when the object is too high to be measured easily.

7.1.1.1. The height of the object can be calculated using the concept of trigonometric ratios

7.1.1.1.1. Light near pool - angle of elevation: 11 degrees - distance: 5.8m

7.1.1.1.2. Clock - angle of elevation: 9 degrees - distance: 2.2m

7.1.1.1.3. Top right corner of vitamin water refrigerator - angle of elevation: 22 degrees - distance: 1.4m