If your solution to stated problem does not match the given solution, you should self-critique per instructions at

   http://vhcc2.vhcc.edu/dsmith/geninfo/labrynth_created_fall_05/levl1_22/levl2_81/file3_259.htm

.

Your solution, attempt at solution.  If you are unable to attempt a solution, give a phrase-by-phrase interpretation of the problem along with a statement of what you do or do not understand about it.  This response should be given, based on the work you did in completing the assignment, before you look at the given solution.

SOLUTIONS/COMMENTARY ON QUERY 8

 ****   Query   problem 7.1.B-10  c = 10 , alpha = 40 deg, right triangle

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                                                  10:55:51

Given Solution: 

 ** c is the hypotenuse.  The angle a opposite the angle alpha satisfies

hypotenuse * sin(alpha) = a so that

a = 10 * sin(40 deg) = 6.43, approx..

We also have

b = c * cos(alpha) = 10 * cos(40 deg) = 7.66 approx..

The remaining angle of the triangle is beta = 90 deg - alpha = 90 deg - 40 deg = 50 deg. **

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  ****   Query   problem 7.1.B-24 cliff height 100 feet, angle of elevation

25 deg.  Dist of ship from shore.

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Your solution:  

Confidence Assessment:

Given Solution: 

 ** The cliff height forms a leg of a right triangle, oppposite the 25 deg angle.

The distance from ship to shore forms the other leg of the triangle, adjacent to the 25 deg angle.

Cliff height / distance from ship to shore = opposite side / adjacent side = tan(25 deg) so

adjacent side = opposite side / tan(25 deg) = 100 ft / tan(25 deg) =  214.5 ft. **

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  ****   Query   problem 7.1.B-36  guy wire 80 ft long makes an angle of 25

deg with a ground; ht of tower?

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                                                  11:07:20

Your solution:  

Confidence Assessment:

Given Solution: 

 **  The guy wire is the hypotenuse of a right triangle for which the altitude is opposite the 25 degree angle.  Thus we have

altitude = hypotenuse * sin(25 deg) = 80 ft * sin(25 deg) = 33.8 ft. **

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  ****   query  problem 7.1.A-72  length of ladder around corner hall widths

3 ft and 4 ft `theta relative to wall in 4' hall, ladder in contact with

walls

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Your solution:  

Confidence Assessment:

Given Solution: 

 ** In the triangle formed by the ladder in the wider hall, `theta is the angle opposite the 4-foot leg of the triangle.  If the length of the part of the ladder in that hall is c1, then c1 = 4 / sin(`theta).

In the triangle formed in the narrower hall, the 3-foot leg of the triangle is parallel to the sides of the wall in the first hall so by corresponding angles `theta is the angle adjacent to that leg, and if c2 is the hypotenuse of that triangle we have c2 = 3 ft / cos(`theta). 

The length of the ladder is therefore

3 ft / cos(`theta) + 4 ft / sin(`theta) or

3 ft sec(`theta) + 4 ft csc(`theta). **

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 ****   query  problem  7.1.A-78  area of isosceles triangle A = a^2

sin`theta cos`theta, a length of equal side

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Your solution:  

Confidence Assessment:

Given Solution: 

 ** If you divide the triangle by its axis of symmetry you get two congruent right triangles, each with angle `theta opposite the altitude and adjacent to the base.  The side a makes up the hypotenuse of either of these triangles.

The altitude of each is therefore a sin(`theta) and the base is a cos(`theta).  The area of each triangle is thus 1/2 * base * height = 1/2 a sin(`theta) a cos(`theta) = 1/2 a^2 sin(`theta) cos(`theta).

The areas of the two right triangles add up to the area of the isosceles triangle.  This area is therefore

2 ( 1/2 a^2 sin(`theta) cos(`theta) ) = a^2 sin(`theta) cos(`theta). **

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 ****   Query   problems 7.2.12  alpha = 70 deg; `beta = 60 deg, c = 4

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                                                  11:48:52

Your solution:  

Confidence Assessment:

Given Solution: 

 ** GOOD STUDENT SOLUTION:

if alpha = 70 deg; and `beta = 60 deg, then `gamma = 50 deg

alpha + `beta + `gamma = 180 deg

70 deg + 60 deg = `gamma = 180 deg.

`gamma = 180 deg - 130 deg

`gamma = 50 deg.

Now for the sides - knowing what the three angles are and knowing that c = 4,

a = :

sin alpha / a = sin`gamma / c

sin70 deg / a = sin 50 deg / 4

a = 4(sin70 deg) / sin50 deg

a is approx. 4.91

b = :

sin `beta / b = sin `gamma/ c

sin 60 deg / b = sin 50 deg / 4

b = 4(sin 60 deg) / sin 50 deg

b is approx. 4.52  **

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 ****   Query   problems 7.2.28  b = 4, c = 5, `beta = 40 deg

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Your solution:  

Confidence Assessment:

Given Solution: 

 ** sin(`gamma) = .80.

Thus `gamma = arcsin(.80) = 53 deg, approx., or 180 deg - 53 deg = 117 deg. Note that we have to consider both angles because the sine doesn't distinguish between the first and second quadrant, whereas the cosine (which is negative in the second quadrant) would.

If `gamma = 53 deg then alpha would be 87 deg.  In this case the Law of Sines tells us that

a = sin(87 deg) * 4 / sin(40 deg) = 6.2, approx..

If `gamma = 117 deg then alpha would be 23 deg so that

a = sin(23 deg) * 4 / sin(40 deg) = 2.8 or so. 

You should draw both triangles to see that both of these solutions are possible. **

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  ****   specify the unknown sides and angles of your triangle.

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                                                  11:57:48

Your solution:  

Confidence Assessment:

Given Solution: 

 if it is possible to draw the triangle or even if it isn't we can solve for

a. so we can say alpha+beta + gamma=180 deg. so alpha + 40+.80=180 so alpha=

139.2 we can then find the value of a by saying sin 139.2/a= sin 40/4 which

is 4 sin 139.2= a sin 40 deg so a= 4.07 and  alpha=139.2

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                                                  11:57:48

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  ****   Query   problems 7.2.40  line-of-sight angles 15 deg and 35 deg with line directly to shore

points are 3 miles apart .  

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Your solution:  

Confidence Assessment:

Given Solution: 

** First form two right triangles.

The first is from ship to shore to lighthouse A.  Angles are 15 deg, 90 deg and 75 deg.

The second is from ship to shore to lighthouse B.  Angles are 35 deg, 90 deg and 55 deg.

Now form the triangle from ship to lighthouse A to lighthouse B.  Let alpha be the angle formed at the ship.  Then

'alpha = 50deg

'beta = 55deg

'gamma = 75deg

a = 3mi (the separation of the lighthouses).

distance to lighthouse A is the side b:

Law of sines tells us that

sin(50deg)/3 = sin(55deg)/b so

b = 3sin(55deg)/sin(50deg)

b = 3.21 mi.

distance to light house B is side c:

By Law of Sines

c = 3(sin75deg)/sin(50deg)

c = 3.78 mi

distance to shore:

Using first right triangle

Theta = 15

Hypotenuse = distance to light house A = 3.21mi

cos`theta = dist to shore / hypotenuse so

dist to short = hypotenuse * cos(`theta) = 3.21 mi * cos(15 deg) = 3.1 mi.

The same distance would be confirmed by solving the other right triangle. **

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