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Triangles Classifications of Triangles Sum of Angles in triangles Pythagorean Theorem Trig Ratios Area of Triangles.

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Presentation on theme: "Triangles Classifications of Triangles Sum of Angles in triangles Pythagorean Theorem Trig Ratios Area of Triangles."— Presentation transcript:

1 Triangles Classifications of Triangles Sum of Angles in triangles Pythagorean Theorem Trig Ratios Area of Triangles

2 Triangle Review Acute Triangle Obtuse Triangle Right Triangle Equilateral Triangle Isosceles Triangle Scalene Triangle Sum of the angles in a triangle is 180

3 Solve for x and classify the triangles by its sides and angles 3x=45y+7=45 x=15y=38 Isosceles Right Triangle 5x=60 x=12 Equilateral, Acute 5x+55=180 5x=125 X=25 75,50,55 Scalene, Acute 3xy+7 5x 552x 3x 10

4 The Pythagorean Theorem In a right triangle, the square of the length of the hypotenuse is equal to the sum of the squares of the lengths of the legs. a 2 + b 2 =c 2 a b c

5 Example 1: Find the Area Because the triangle is isosceles, the base is bisected. 8m 10m h Use pyth. Thm to find “ h ” a 2 +b 2 =c 2 5 2 +h 2 =8 2 25+h 2 =64 h 2 =39 h=6.24 Area of Triangle A=1/2 bh A=1/2(10)(6.24) A=31.2m 2 =6.24 55

6 Pitcher ’ s Mound 65ft Home Plate 1st base 2nd base 3rd base 50ft Example 2: In slow pitch softball, the distance between consecutive bases is 65 ft. The pitcher ’ s plate is located on the line between second based and home plate 50 ft from home plate. How far is the pitcher ’ s plate from second base? Justify your answer You can use the Pyth. Thm: a 2 +b 2 =c 2 65 2 +65 2 = x 2 4225+4225=x 2 8450 = x 2 91.9ft from home plate to 2nd base = x 91.9ft Total distance - PM to HP = 2nd to PM 91.9 - 50 = 41.9ft

7 Pythagorean Triple: 3 positive integers a,b,c, that satisfy a 2 +b 2 =c 2 Example: 3,4,5 represent a Pythagorean Triple 3 2 +4 2 =5 2 9+16 = 25 25=25

8 Common Pyth Tripples

9 Trig Ratios: B C A a c b Sin A = Opposite Side Hypotenuse Cos A = Adjacent Side Hypotenuse Tan A = Opposite Side Adjacent Side Opposite Hypotenuse Adjacent

10 Example 1: Find the ratio of the sin A, cos A and Tan A Sin A = Opp Hyp. Opposite B C A 3 5 4 Hypotenuse Adjacent = 3 5 Cos A = Adj Hyp = 4 5 Tan A = Opp Adj = 3 4

11 Example 2: Find the ratio of the sin B, cos B and Tan B Sin B = Opp Hyp. Adjacent B C A 3 5 4 Hypotenuse Opposite = 4 5 Cos B = Adj Hyp = 3 5 Tan B = Opp Adj = 4 3

12 Sin, Cos and Tan on your Calculator Use your calculator: Cos 13º = _______.9744 Sin 27º = _______.4540 Tan 66º = _______ 2.2460

13 Example 4: Find the height of the silo. 48º 100 ft x You can use Tan Ratio: Tan A = opp adj Tan 48 = x 100 S o l v e b y c r o s s m u l t. X = 100 ● tan 48 X = 111 ft.

14 Example 5: You are measuring the height of a tower. You stand 154 ft. from the base of the tower. You measure the angle of elevation from a point on the ground to the top of the tower to be 38º. Estimate the height of the tower. 38º 154 ft x Tan A = opp adj Tan 38 = x 154 X = 154 ● tan 38 X = 120 ft.

15 Example 6: Other Variations: Solve for x 14 22 x 479x 35 15 x Sin 22 = x/14 14Sin22 = x Cos 47 = 9/x xCos 47 = 9 X = 9/cos47 Sin 35 = 15/x xSin35 = 15 X = 15/Sin35 52 19 x Cos 52 = x/19 19Cos 52 = x

16 Example 7: Solve the Right Triangle A B C 8 10 x Sides: AB = 8 BC = 10 Missing AC: To find AC use pyth thm 8 2 +10 2 =x 2 64+100 = x 2 164 = x 2 12.8 = x, AC = 12.8

17 Example 7: Solve the Right Triangle A B C 8 10 x Angles: <B = 90 Missing <A and <C: To find find the missing angles, we will use INVERSE trig functions. Tan A = opp adj Tan A = 10 8 Tan A = 1.25 To get A by itself, we must do the opposite of Tan. This is called INVERSE TAN, it is Tan -1 on your calculator Tan -1 Tan A = Tan -1 1.25 A = 51.34º opp adj

18 Example 7: Solve the Right Triangle A B C 8 10 x Angles: <B = 90 <A = 51.34 Missing <C: To find find the missing angles, we will use INVERSE trig functions. Tan C = opp adj Tan C = 8 10 Tan C =.8 To get A by itself, we must do the opposite of Tan. This is called INVERSE TAN, it is Tan -1 on your calculator Tan -1 Tan C = Tan -1.8 C = 38.66 opp adj

19 Example 7: Solve the Right Triangle A B C 8 10 x Angles: m<B = 90 m<A = 51.34 m<C = 38.66 Sides: AB = 8 BC = 10 AC = 12.8

20 Example 8: Find the area of each triangle 14 10 50 1812 15


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