SECONDARY MATH Area of a Triangle and Law of Sines

Transcription

1 SECONDARY MATH Area of a Triangle and Law of Sines

2 Goal: Be the first team to find (r j h g f)(x). WARM UP COMPOSITION OF FUNCTIONS Person #1 f(x) = x 2 7x + 6 Person #2 g(x) = Person #3 h(x) = Person #4 j(x) = 3log(x + 11) Person #5 r(x) = 4 x+2 3x + 1 Start with f(3) Start with f(-2)

3 TRIGONOMETRY SOH CAH TOA = Inverses &= =!" # '&=!" # $%=!" # (=!" #

4 SOLVE THE TRIANGLE USING TRIG RATIOS What are the measurements of the missing side and the two other angles? Use trig ratios to solve. Round to the nearest hundredth Make sure calculators are set to degrees (Mode)

5 A rea = 1 2 AREA OF A TRIANGLE *base*height What is the area of the triangle? A rea = 1 2 (4)(3) = 6 square units

6 A rea = 1 2 *base*height YOUR TURN What is the area of the triangle? A rea = 1 2 (12)(9) = 54 square units

7 AREA OF A TRIANGLE What if the triangle isn t a right triangle? A º 8 C 52.2º 48º B 10

8 IT IS POSSIBLE TO FIND THE AREA OF A TRIANGLE USING TRIGONOMETRY WHEN GIVEN TWO SIDES AND THE INCLUDED ANGLE. Label your triangle by labeling each side with lower case a, b, and c and then label the opposite angle of each side with the same letter except upper case. Start with any triangle that has angles A, B, and C and side lengths a, b, and c, where a is the side opposite angle A, b is the side opposite angle B, and c is the side opposite angle C.

9 First, construct an altitude (height), h, from the vertex of one of the angles to the side opposite the angle. The two triangles formed, ABD and BDC, are right triangles. Now find the measure of h in terms of A and side c using the sine ratio for A. )= # Now we solve for h: &)=h

10 The base of the entire triangle is b and the height is h. We can now substitute &)=hinto the area formula for h, the height. )= 1 2 (,%')(h'.h() )= 1 2, &) Note: This formula works for acute, obtuse, and right triangles. )= 1 2,&())

11 AREA OF A TRIANGLE GIVEN TWO SIDES AND THE INCLUDED ANGLE )/'%= 1 2,&()) )/'%= 1 2 %&(1) )/'%= 1 2 %,()

12 Find the area of the triangle. AREA OF A TRIANGLE )/'%= 1 2,&()) A )/'%= 1 2 %&(1) º 8 )/'%= 1 2 %,() C 52.2º 48º B 10 Area = 29.7 square units

13 MORE EXAMPLES - Find the area of a triangle with sides a = 11, b = 5 and C = 20. Find the area of a triangle with sides b = 13, c = 7, and A = 43.

14 MORE EXAMPLES - Find the area of a triangle with sides b = 11, c = 24 and A = 76. Find the area of a triangle with sides a = 10, c = 22, and B = 101.

15 You have used sine, cosine, and tangent to solve triangles. These ratios can only be used to solve right triangles. What do we do if the triangle is not a right triangle? C b a A c B

16 LAW OF SINES ) % =1, = & % ) =, 1 = &

17 PROOF OF THE LAW OF SINES Start with and triangles that has angles A B, and C and side lengths, a, b, and c, where a is the side opposite angle A, etc. Construct an altitude, h, from one of the angles to the side opposite the angle. The two triangles formed ABD and BDC, are right triangles. Use the definition of sine to relate the base angles, A and C, to the hypotenuse of each and the altitude )= # =

18 Use multiplication property to solve for h )= # &)=h = % =h NOW WE JUST HAVE TO DO SOME ALGEBRA Use the transitive property of equality to set the equations equal to one another &) =% Finally, use the division property of equality. ) % = &

19 a = , c = , B = 40 First check to see if you can find any other values without using the Law of Sines. B = 180 ( ) = 40 You then need to decide what is given so you know what ratios in the Law of Sines to use. (We have B and b) EXAMPLE USE THE LAW OF SINES TO SOLVE THE TRIANGLE Use = 4 to solve for a Plug in the values given: = Solve for a: %40 = %= Now solve for c by plugging values into the Law of Sines. & 22 = &40 = 2422 &=

20 B = 48, b = 5.773, and c = EXAMPLE #2 Solve a triangle where A = 70, C = 62, and a = 7.3. Draw the triangle if needed.

21 A= 61, a = , and b = EXAMPLE #3 Solve a triangle where B = 16, C = 103, and c = 12. Draw the triangle if needed.

22 A = , C = , and c = EXAMPLE #4 Solve a triangle where B = 103, b = 61, and c = 46

23 THE AMBIGUOUS CASE (SSA) If you are given two angles and one side (ASA or AAS) or SAS, the Law of Sines will easily provide ONE solution for a missing side. However, the Law of Sines has a problem dealing with SSA. If you are given two sides and one angle, where you must find an angle, the Law of Sines could possibly provide you with one or more solutions or even no solution at all. Notice the two different angles C could be and the two different angles B could be.

24 Solve for angle B: 2 8 A = = 31 1= = 3 EXAMPLE #5 USE THE LAW OF SINES TO SOLVE THE TRIANGLE Now we need to test to see if more than one triangle exists: (180 found angle) + given angle = If sum > 180 there is only one triangle If sum < 180 there are two triangles ( ) + 40 = > 180 There will be only one triangle for the given info. Can we make it easier? Now find C. 180 ( ) = Now find the missing side c = & &40 = &= B = , C = , c = 4.242

25 2 A Solve for angle B: = 25 B = 61 1= = Now we need to determine if there will be two triangles. EXAMPLE #6 USE THE LAW OF SINES TO SOLVE THE TRIANGLE. Now find C. 180 ( ) = Now find the missing side c = & &35 = & = ( ) + 35 = < 180 There are two triangles for the given info. B = , C = , c =

26 Notice that this new triangle has the same given information. )1 7 +B1 7 =180(they form a straight angle) WE CAN DRAW THE SECOND TRIANGLE BY SWINGING BC TOWARDS AFORMING AN ISOSCELES TRIANGLE AND CREATING TRIANGLE A(B2)C Now find the missing side ) = ) )1 7 35= )1 7 = )1 7 = = B = , C = , c = )1 7 = =14.886

27 EXAMPLE #7 Solve a triangle where A = 150, a = 9.3 and b = 41. No Solution

28 Triangle #1: A = , C = , c= Triangle #2: A = , C = , c=5.395 EXAMPLE #8 Solve the triangle where B = 60, a = 18, b = 16.