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- 1. Right Triangle Problems, Law of Sines, Law of Cosines & Problem Solving T- 1-855-694-8886 Email- info@iTutor.com By iTutor.com
- 2. Right Triangle Problems Any situation that includes a right triangle, becomes solvable with trigonometry. Angle of Elevation Angle of Depression © iTutor. 2000-2013. All Rights Reserved
- 3. This diagram has an inscribed triangle whose hypotenuse is also the diameter of the circle.A B C Given: = 40o, AB = 6 cm Determine the area of the shaded region. Solution: Sin 40o = 6/AC; AC = 6/ sin40o = 9.33 cm AC/2 = radius of the circle = 4.67 cm; BC = 6/tan 40o =7.15 cm Ao = r2 = 68.4 cm2, A = ½bh = 21.5 cm2; A = 46.9 cm2 © iTutor. 2000-2013. All Rights Reserved
- 4. This diagram has a right triangle within a cone that can be used to solve for the surface area and volume of the cone.h r l Given: = 55o, l = 6 cm Determine the area and volume of the cone. Solution: Sin 55o = h/l; h = 6 x sin55o = 4.91 cm tan 55o = h/r; r = h/ tan 55o = 4.91/ tan 55o = 3.44 cm Acone = r(r + l) = (3.44)(3.44 + 6)cm2; Acone = 102 cm2 Vcone =⅓ r2h = ⅓ (3.44)2(4.91)cm3; Vcone = 60.8 cm3 © iTutor. 2000-2013. All Rights Reserved
- 5. The angle of elevation of a ship at sea level to a neighboring lighthouse is 2o . The captain knows that the top of lighthouse is 165 ft above sea level. How far is the boat from the lighthouse? h x Solution: tan 2o = h/x; 1 mile = 5280 ft, so the ship is .89 mile away from the lighthouse. x = h/tan 2o x = 165/ tan 2o = 4725 ft; © iTutor. 2000-2013. All Rights Reserved
- 6. Function Values – Unit Circle The unit circle is a circle with a radius of 1 unit. x y Sin = y; Cos = x; Tan = y/x The coordinate of P will lead to the value of x and y which in turn leads to the values for sine, cosine, and tangent. Use the reference angle in each quadrant and the coordinates to solve for the function value. © iTutor. 2000-2013. All Rights Reserved
- 7. The reference angle is always measured in its quadrant from the x – axis. Quadrant I – Angle = Quadrant II – Angle = (180 - ) Quadrant III – Angle = (180 + ) Quadrant IV – Angle = (360 - ) Quadrant I – P (x, y) Quadrant II – P (-x, y) Quadrant III – P (-x, -y) Quadrant IV – P (x, -y) With the values of changing from (+) to (-) in each quadrant, and with the functions of sine, cosine, and tangent valued with ratios of x and y, the functions will also have the sign values of the variables in the quadrants. © iTutor. 2000-2013. All Rights Reserved
- 8. Degrees 0 30 45 60 90 120 135 150 Sine 0 ½ 2/2 3/2 1 3/2 2/2 ½ Degrees 180 210 225 240 270 300 315 330 Sine 0 -½ - 2/2 - 3/2 -1 - 3/2 - 2/2 -½ Cosine 1 3/2 2/2 ½ 0 -½ - 2/2 - 3/2 Tangent 0 3/3 1 3 - - 3 -1 - 3/3 Cosine -1 - 3/2 - 2/2 -½ 0 ½ 2/2 3/2 Tangent 0 3/3 1 3 - - 3 -1 - 3/3 © iTutor. 2000-2013. All Rights Reserved
- 9. Example: Cos 240 = _______ 60 P (-½, - 3/2) - ½ Example: Sin 240 = _______ Example: Tan 240 = _______ 3 - 3/2 240 © iTutor. 2000-2013. All Rights Reserved
- 10. Law of Sines Law of Sines – For a triangle with angles A, B, C and sides of lengths of a, b, c the ratio of the sine of each angle and its opposite side will be equal. Sin A = Sin B = Sin C a b c A B C c b a h Proof: Sin A = h/b; Sin B = h/a h = b Sin A, h = a Sin B b Sin A = a Sin B; Sin A = Sin B a b © iTutor. 2000-2013. All Rights Reserved
- 11. Proof, continued: A B C c b a kk is the height of the same triangle from vertex A Sin C = k/b and Sin B = k/c k = b Sin C and k = c Sin B b Sin C = c Sin B; Sin B = Sin C b b c Conclusion: Sin A = Sin B = Sin C a b c © iTutor. 2000-2013. All Rights Reserved
- 12. Law of Sines Let’s take a closer look: Suppose A < 90o A c a If a = c Sin A, one solution exists a A c c Sin A If a < c Sin A, no solution exists © iTutor. 2000-2013. All Rights Reserved Let’s take a closer look: Suppose A < 90o A ac If a > c Sin A and a c, one solution exists c Sin A a A c c Sin A If c Sin A < a < c , two solution exists a
- 13. Let’s take a closer look: Suppose A 90o A a c If a < c, no solution exists A a If a > c , one solution exists c
- 14. Law of Cosines A B CD b a c a -x x h b2 = h2 + x2; h2 = b2 - x2 Cos C = x/b ; x = b Cos C c2 = h2 + (a-x)2; c2 = h2 + a2 –2ax + x2 Substitute and we get: c2 = (b2 - x2)+ a2 –2a(bCos C) + x2 c2 = b2 + a2 – 2abCos C For any triangle given two sides and an included angle, a2 = b2 + c2 – 2bcCos A b2 = a2 + c2 – 2acCos B c2 = b2 + a2 – 2abCos C PROOF: © iTutor. 2000-2013. All Rights Reserved
- 15. Law of Cosines - example Given: A = 50o; AB = 8 cm, AC = 14 cm x 2 = 8 2+ 14 2 – (8)(14)Cos50 Find x. A B C x x 2 = 260 – 72; x 2 = 188 x = 13.7 cm © iTutor. 2000-2013. All Rights Reserved
- 16. Law of Cosines – Hero’s Formula For any triangle, the area of the triangle can be determined with Hero’s Formula. s = ½(a + b + c) where s = semi-perimeter of the triangle. A = s(s – a)(s – b)(s – c) Given: ABC; a = 7, b = 24, c = 25 A = 28(28 – 7)(28 – 24)(28 – 25) = 84 cm 2 7cm 25 cm 24 cm S = ½ (7 + 24 + 25) = 28 Check: A = ½(b)(h) = ½(7)(24) = 84 cm 2 Example © iTutor. 2000-2013. All Rights Reserved
- 17. The End Call us for more information: www.iTutor.com 1-855-694-8886 Visit

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