NO LINKS!!! Draw and solve each triangle. 1. a= 3, b= 4, C= 40°
2. a= 3, c= 2, B=110°
3. a= 12, b= 13, , c= 5
4. a= 5, b= 8, c = 9

The missing values are highlighted in bold, which are approximate to four decimal places (with the exception of B = 90 in problem 3).

Lowercase letters a,b,c are side lengths; Uppercase letters A,B,C are angles opposite the sides mentioned.

The drawings are shown below.

==========================================================

Explanation:

Problem 1

Use the Law of Cosines to solve for side c

c^2 = a^2 + b^2 - 2*a*b*cos(C)

c^2 = 3^2 + 4^2 - 2*3*4*cos(40)

c^2 = 6.61493337

c = sqrt(6.61493337)

c = 2.57195128

c = 2.5720

Now apply the Law of Sines to find angle A

sin(A)/a = sin(C)/c

sin(A)/3 = sin(40)/2.57195128

sin(A) = 3*sin(40)/2.57195128

sin(A) = 0.74976647

A = arcsin(0.74976647)

A = 48.57015285

A = 48.5702

Once we know two angles of a triangle, we can then subtract their sum from 180 to get the third missing angle.

A+B+C = 180

B = 180 - (A+C)

B = 180 - (48.57015285+40)

B = 91.42984715

B = 91.4298

-----------------------------------

Problem 2

Like with the previous problem, we can use the Law of Cosines to solve for the missing side.

b^2 = a^2 + c^2 - 2*a*c*cos(B)

b^2 = 3^2 + 2^2 - 2*3*2*cos(110)

b^2 = 17.10424172

b = sqrt(17.10424172)

b = 4.13572747

b = 4.1357

Now use the Law of Sines to find angle A

sin(A)/a = sin(B)/b

sin(A)/3 = sin(110)/4.13572747

sin(A) = 3*sin(110)/4.13572747

sin(A) = 0.68164014

A = arcsin(0.68164014)

A = 42.9719427

A = 42.9719

Then we can wrap things up like so

A+B+C = 180

C = 180 - (A+B)

C = 180 - (42.9719427+110)

C = 27.0280573

C = 27.0281

-----------------------------------

Problem 3

Use the Law of Cosines to find angle A.

a^2 = b^2 + c^2 - 2*b*c*cos(A)

12^2 = 13^2 + 5^2 - 2*13*5*cos(A)

144 = 194 - 130*cos(A)

-130*cos(A) = 144 - 194

-130*cos(A) = -50

cos(A) = (-50)/(- 130)

cos(A) = 0.38461538

A = arccos(0.38461538)

A = 67.38013505

A = 67.3801

We could use the Law of Cosines to find angle B, but I'll use the Law of Sines instead.

sin(A)/a = sin(B)/b

sin(B) = b*sin(A)/a

sin(B) = 13*sin(67.38013505)/12

sin(B) = 0.99999999998574

sin(B) = 1

B = arcsin(1)

B = 90

It turns out we have a 5-12-13 right triangle (though usually c is the longest side instead of b).

Now we could use the Law of Cosines to find angle C, but the steps shown below are much quicker in my opinion that don't involve any trig functions.

A+B+C = 180

C = 180 - (A+B)

C = 180 - (67.38013505 + 90)

C = 22.61986495

C = 22.6199

-----------------------------------

Problem 4

The steps will basically be identical to problem 3, but with different numbers of course.

First we need the Law of Cosines to find angle A.

a^2 = b^2 + c^2 - 2*b*c*cos(A)

5^2 = 8^2 + 9^2 - 2*8*9*cos(A)

25 = 145 - 144*cos(A)

-144*cos(A) = 25 - 145

-144*cos(A) = -120

cos(A) = (-120)/(-144)

cos(A) = 0.83333333

A = arccos(0.83333333)

A = 33.55730976

A = 33.5573

Now use the Law of Sines to find angle B.

sin(A)/a = sin(B)/b

sin(B) = b*sin(A)/a

sin(B) = 8*sin(33.55730976)/5

sin(B) = 0.88443328

B = arcsin(0.88443328)

B = 62.18186103

B = 62.1819

Lastly,

A+B+C = 180

C = 180 - (A+B)

C = 180 - (33.55730976 + 62.18186103)

C = 84.26082921

C = 84.2608

-----------------------------------

Side notes:

$450.00 for 6 weeks

or

$300

step-by-step explanation:

$200.00 for first week $50.00 x5 equals $250 plus $200 equals $450.00 for six weeks.

$200 for first week

$200 x 0.10 = $20

$20 x 5 = $100

$200 + $100 = $300

50*10 is greater

step-by-step explanation:

if you add the zero from 10 to 50, it will give you a bigger number.