# 1)Given a = 11, b = 6, and c = 8, use the Law of Cosines to solve the triangle for the value of C. Round answer to two decimal places. 2)Given C = 112 degrees, a = 13, and b = 9, use the Law of...

1)Given *a* = 11, *b* = 6, and *c *= 8, use the Law of Cosines to solve the triangle for the value of *C*. Round answer to two decimal places.

2)Given *C *= 112 degrees, *a* = 13, and *b* = 9, use the Law of Cosines to solve the triangle for the value of *c*. Round answer to two decimal places.

3)Given *a* = 6.7, *b *= 9.8, and *c *= 7.88, use Heron's Area Formula to find the area of triangle *ABC*. Round answer to two decimal places.

4)A straight road makes an angle, *A,* of 11 degrees with the horizontal. When the angle of elevation, *B*, of the sun is 57 degrees, a vertical pole beside the road casts a shadow 7 feet long parallel to the road. Approximate the length of the pole. Round answer to two decimal places.

5)After a severe storm, three sisters, April, May, and June, stood on their front porch and noticed that the tree in their front yard was leaning 7o from vertical toward the house. From the porch, which is 104 feet away from the base of the tree, they noticed that the angle of elevation to the top of the tree was 31o. Approximate the height of the tree. Round answer to two decimal places.

6)A vertical pole 27 feet tall stands on a hillside that makes an angle of 19o with the horizontal. Determine the approximate length of cable that would be needed to reach from the top of the pole to a point 80 feet downhill from the base of the pole. Round answer to two decimal places.

7)A vertical pole 32 feet tall stands on a hillside that makes an angle of 17o with the horizontal. Determine the approximate length of cable that would be needed to reach from the top of the pole to a point 60 feet downhill from the base of the pole. Round answer to two decimal places.

8)A triangular parcel of land has sides of lengths 450, 400 and 650 feet. Approximate the area of the land. Round answer to nearest foot.

9)In the figure below a=6, b=12, and c=13. Use this information to solve the parallelogram for `theta` . The diagonals of this parallelogram represent by c and d.

10) In the figure below a=6, b=12, and c=10. Use this information to solve the parallelogram for ``. The diagonals of this parallelogram represent by c and d.

11) Given *A *= 55o, *B* = 63o, and *a* = 7.1, use the Law of Sines to solve the triangle for the value of *b*. Round answer to two decimal places.

12)Determine a value for *b* such that a triangle with *A* =61o and *a* = 9 has only one solution.

*print*Print*list*Cite

### 1 Answer

1) The Law of Cosines uses the formula: `a^2= b^2+c^2-2bcCosA` Note that the side length to be solved always corresponds to the specified angle (for example, side a solves with cos A, side b solves with cos B and so on).` `

We are looking for angle C: `therefore c^2=a^2+b^2-2ab cos C`

`therefore 8^2=11^2+6^2-2(11 times 6) cos C`

`therefore 64=121+36-132 cos C`

`therefore 64-121-36= -132cosC`

`therefore 0.704545=cos C` Note: It is easier to substitute and manipulate the formula this way rather than to memorize 2 formulas).

`therefore C=45.21 degrees`

2) Use the same formula for the Law of Cosines and find a side length. We are looking for side c:

`therefore c^2=9^2+13^2 - 2(9 times13)cos 112`

`therefore c^2=250-234cos112`

`therefore c^2=162.34`

`therefore c=sqrt162.34`

`therefore c=12.74 units`

3) Heron's Area Formula uses half of the triangle's perimeter and the formula is therefore two-fold:

`s=(a+b+c)/2`

`A=sqrt(s(s-a)(s-b)(s-c))`

a,b and c relate to the side lengths:

`therefore s=6.7+9.8+7.88`

`s=12.19`

`A=sqrt(12.19(12.19-6.7)(12.19-9.8)(12.19-7.88))`

`A=sqrt(689.3682)`

`A=26.26 units^2`

4) There are various calculations necessary to calculate this:

i) The angle required meets the road and therefore we need 57 degrees - 11 degrees = 46 degrees

ii) As it casts a shadow of 7 ft, this is the length running along the road

iii) The pole obviously sits at 90 degrees to the road and 90-11=79 degrees

iv) In the created triangle, the angle will be 180- 79= 101 degrees (which has formed as the shadow is cast down the road)

v) This leaves the angle opposite the 7 ft shadow as 33 degrees(angles of a triangle = 180 degrees)

Now that we have sufficient information, we can apply the Law of Sines:

`a/(sinA)=b/sinB` Let a = length of the pole. Therefore:

`a/sin 46=6/sin33`

`therefore a=7.92 units` Therefore the pole length is 7.92 units.

Please re-post your other questions separately as there are just too many for one question. I recommend possibly 4 per post. Thank you.

**Ans:**

** 1) Angle C = 45.21 degrees**

**2) c=12.74 units**

**3)Area=26.26 units ^2**

**4)Pole = 7.92 units**