Dear class I am happy to be chosen as the scribe for today’s class.It good that we are back we the good habit of posting that was about to be stopped because it have been almost one week that nobody have posted. Today Dr AVIATAR gave us a pretest which seemed to be hard at the beginning although we were pretending to do it as if it was a test. The pretest took 30 minutes and after we tried to do the question in group. later she tried to explain the question.
What I want to remind everybody that the test is for tomorrow Wednesday 09 October 2008 and we have to get prepared seriously
There are couple of things that i want to remind every body about tomorrow test
1 if you are given a any problem to solve, problem which has already a drawing and in some of the cases that we are dealing with in this unit relate with triangle
- when given some sides and angles of the triangle
we use the sin law to solve the equation
first s
Find angle B.
Angle B = 180o - (43o + 57o) = 80o
Now, we use the law of sines to
find the other sides lengths.
c a
----- = -----
sin C sin A
a(sin C)
c = --------
sin A
Plug in any known information.
4.56(sin 570)
c = -------------
sin 43o
Use a calculator to find the sines.
4.56(.8387)
c = -----------
.6820
c = 5.61
Now solve for b.
b a
----- = -----
sin B sin A
a(sin B)
b = --------
sin A
Plug in any known information.
4.56(sin 80o)
b = -------------
.6820
Use a calculator to find the sines.
4.56(.9848)
b = -----------
.6820
b = 6.58
There is another case when you can use the law of sines. When you know two sides and an angle opposite one of the sides, the law of sines can be used. However, with this case, you have to be aware that there might not be a solution, or there may be two! One solution is also possible. Example:
2. Problem: In triangle ABC, a = 15, b = 25, and angle A = 47o.
Solve the triangle.
Solution: Start out by looking for the measure of angle B
a b
----- = -----
sin A sin B
b(sin A)
sin B = --------
a
Plug in any known information.
25(sin 47o)
sin B = -----------
15
Use a calculator to find the sine.
25(.7314)
sin B = ---------
15
sin B = 1.219
Since an angle cannot have a sine greater than 1, there is no solution for this triangle
.
- When asked to verify number of triangle that can be made from different angles and sides given
- When given a trigonometric equation (Trigonometric Equations are equations that contain expressions such as sin x. You solve a trigonometric equation in the same way as any other equation.) the most important thing is to start by isolating the term then apply the sin law or cosine law
Solve for x: 2sin x = 1.
Solution: Solve for sin x by
dividing by 2.
sin x = (1/2)
The circle below can show the range and domain of sin and cosine of and angle
thank you for reading the blog please prepare your test and good luck
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