Law of sine and cosines
Upcoming SlideShare
Loading in...5
×
 

Law of sine and cosines

on

  • 849 views

 

Statistics

Views

Total Views
849
Views on SlideShare
849
Embed Views
0

Actions

Likes
0
Downloads
52
Comments
0

0 Embeds 0

No embeds

Accessibility

Upload Details

Uploaded via as Microsoft PowerPoint

Usage Rights

© All Rights Reserved

Report content

Flagged as inappropriate Flag as inappropriate
Flag as inappropriate

Select your reason for flagging this presentation as inappropriate.

Cancel
  • Full Name Full Name Comment goes here.
    Are you sure you want to
    Your message goes here
    Processing…
Post Comment
Edit your comment

Law of sine and cosines Law of sine and cosines Presentation Transcript

  • Right Triangle Problems, Law of Sines, Law of Cosines & Problem Solving T- 1-855-694-8886 Email- info@iTutor.com By iTutor.com
  • 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
  • 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 View slide
  • 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 View slide
  • 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
  • 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
  • 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
  • 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
  • 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
  • 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
  • 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
  • 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
  • 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
  • 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
  • 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
  • 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
  • The End Call us for more information: www.iTutor.com 1-855-694-8886 Visit