Aim: Using Appropriate Formulas Course: Alg. 2 & Trig. Aim: What to do, What to do?!? So many...

Aim: Using Appropriate Formulas Course: Alg. 2 & Trig.

Aim: What to do, What to do?!? So many formulas!! Where do we begin?

Do Now:

Regents Question

How many distinct triangles can be forms if A = 30o, side b = 12 and a = 8?

1) 1 2) 2 3) 3 4) 0


x1 2 3 4 5 6-5 -4 -3 -2 -1 0

i

2i

3i

4i

5i

-4i

-3i

-2i

-i

-5i

-6i

yi

Adding Complex Numbers Graphically

(2 + 3i)

(2 + 3i) + (3 + 0i)

(3 + 0i)

(5 + 3i)

= (2 + 3) + (3i + 0i) = = 5 + 3i

vector: 2 + 3i

vector: 3 + 0i

vector: 5 + 3i


Adding Vectors

Vector - a directed line segment that represents directed force notation: OS

R

The vectors that represent the applied forcesform two adjacent sides of a parallelogram, and the vector that represents the resultantforce is the diagonal of this parallelogram.

O

P S

resultant force


The Laws!

a2 b2 c 2 2bc cos ALaw of Cosines:

c

sinC

b

sinB

a

sinALaw of Sines:

Law of Cosines:

cos A b2 c2 a2

2bc

A = 1/2 ab sinCArea of Triangle:


General Guidelines

c

sinC

b

sinB

a

sinALaw of Sines:

Use the Law of Sines when the known information involves ASA, AAS, or SSA.

a2 b2 c 2 2bc cos ALaw of Cosines:

Law of Cosines:

cos A b2 c2 a2

2bc

Use the Law of Cosines when the known information involves SAS or SSS.


Model Problem

To determine the distance between 2 points, A and B, on opposite sides of a swampy region, a surveyor chose a point C that was 350 meters from point A and 400 meters from point B. If the measure of ACB was found to be 105º40’, find to the nearest meter, the distance, AB, across the swampy region.

AB2 AC2 BC2 2( AC)(BC)cosC

Given 2 sides & included angle:

Substitute and solve:

AB2 282500 280000( .2700)

A B

C

400350 105º40’

AB2 3502 4002 2(350)(400)( .2700)

AB2 = 358100 AB = 598.414AB = 598 to nearest meter

Draw:

Law of Cosines


Model Problem

A surveyor on the ground takes two readings of the angle of elevation of the top of a tower. From 150’ apart, the measures are 50o and 70o. Find the tower’s height to the nearest foot.

150’

70o50o

h

B D

A

C

Find AD in ΔABD using Law of Sines;

then work in ΔADC for find AC.

AC 316’


Model Problem

PA and PB are tangent to circle O at points A and B respectively. If PA = 10 cm and mP = 34o, find the length of chord AB to the nearest centimeter.

B

P

A10

34o

Tangents to a circle from an external point are congruent, making PB = 10.

With SAS known, use Law of Cosines.

AB 6 cm.


Model Problem

A canoe race is to be run over a triangular course marked by buoys A, B, and C. The distance between A and B is 100 yards, that between B and C is 160 yards, and that between C and A is 220 yards. Find to the nearest degree, the mABC.


Model Problem

A diagonal of a parallelogram is 50 centimeters long and makes angles of 37o 10’ and 49o 20’, respectively, with the sides. Find the length of the shorter side of the parallelogram to the nearest centimeter.


Regents Question – 4 points

In triangle ABC, mA = 40 and mB = 56. The longest side of the triangle is 36 cm. Find the length of the shortest side to the nearest tenth of a centimeter.


Model Problem

A vertical transmitting tower AB is located on a slope that is inclined 15o to the horizontal. At a point C, 80 feet down the slope from the foot of the tower, the tower subtends an angle of 40o 10’. Find to the nearest foot the height of the tower.

15o

h

80’

40o 10’


Regents Question

In triangle DEF, side e = 10, f = 8 and mD = 110. Find the length of the third side to the nearest tenth.

1) 218.7 2) 109.3 3) 14.8 4) 10.5


Model Problem

A surveyor at point P sights two points X and Y that are on opposite sides of a lake. If P is 200 m. from X and 350 m. from Y, and mXPY = 40, find the distance from X to Y to the nearest meter.


Model Problem

Some nylon fabric will be cut to cover thekite frame shown below. Diagonal AC is29 inches. What size should the angles beat A, B, C, and D?

B

D

C

A 16 in.

26 in.

16 in.

26 in.


Model Problem

Two forces act on a body at an angle of 72o, resulting in a force whose magnitude is 110 lb. If the magnitude of one of the original forces is 80 lb., find the magnitude of the other to the nearest pound.

110 lb.

80 lb

.

108o72o

Draw a parallelogram of forces.

A B

CD

Opposite sides are congruent: AD = BC = 80

Consecutive angles are supplementary: mABC = 108o

With SSA known in ΔABC, apply Law of Sines. To find AB, you must know mACB.

80 lb

.

54 lb.

?


Regents Questions – 6 points

The magnitude of the resultant of two forces acting on a body is 90 lbs. The angles between the forces and the resultant are 22030’ and 56o45’. Find the magnitude of the larger force to the nearest tenth of a pound.

Aim: Using Appropriate Formulas Course: Alg. 2 & Trig. Aim: What to do, What to do?!? So many...

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