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...

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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**