6.1 – Law of Sines
Jan 10th, 2011 by corricelli
Hello,
Let’s use this section to post questions/ respond to questions in Chapter 6, Section 1.
Thanks,
Mrs. Corricelli
12 Responses to “6.1 – Law of Sines”
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For the SSA triangles I’m a little confused on why 27 has no solutions but 29 has two, a read the rules and understand why a must be larger then the height on an acute triangle, however both 27 and 29 are acute and side a is smaller then side b so I’m confused on what the difference is.
Hi, can anyone provide any guidance for # 51…i got the same diagram they have on calc chat, but I am having trouble putting it all together for b. Thanks.
27 has no solutions because if you actually do it out, sinB = 1.0781…and sin has to be inbetween -1 and 1…
i don’t understand how to get all the angle measures for #49. i understand how to get angle A = 46 degrees..but i don’t get the others..can anyone help?
Can anyone tell me what the homework for tonight, 1/11/11, is?
I’m a little confused on how to go about solving 37. In the example thy give you one more
Sorry I accidentally sent that unfinished post, but to continue,
In the example there is one more piece of information than in the problem. Can anyone who noticed this also give some input? Or someone who has solved the problem give me a starting point?
@callum hw is 6.1 #7,23,27,29,37,47,51,55
@sarah i haven’t checked calcchat, but you can make right triangles to find other angles, eventually getting enough information to find the “dock” angle. you can extend the line of the 28 degrees (near the tree) and the 41 degrees (the gazebo) so that they match horizontally with the dock point, creating 2 right triangles. there, you can find the missing angles of those triangles and take that total and subtract it by 180 (because they all share a common vertex( the gazebo)) to get the angle you need for the law of sines.
@haley yeah, i’m confused by that too
Haley and Yuyu,
Sorry to get back to you ladies so late. I was working on another project here and lost track of time. Anyhow, on 37, I would look at the table on p. 430. Draw the triangle (note A is acute) and then strategically pick b (the missing side, as you put it Haley) so that it has one solution (so a has to be greater than or equal to b), two solutions (a is between h and b), and no solution (a is less than h). The pictures really help! Once you know a is acute (since it is 10 degrees) then just use the table.
Thanks Yuyu for giving Callum the hw assignment!
Keep me posted!
Mrs. Corricelli
Sarah,
Remember the geometry rule about exterior angles of any triangle, also! The exterior angle of any triangle is equal to the sum of the nonadjacent interior angles. See this site: http://www.regentsprep.org/regents/math/geometry/GP5/LExtAng.htm for more detail.
Happy Tuesday!
Mrs. Corricelli
Sam,
When it comes to applying the rules, you really need h to make a judgement call and/or to apply these rules. You can get it using the derivations in class. You can assume sinA=h/b so h=bsinA, if the triangle is set up correctly. I prefer to use the fact that the sine value is outside of sine’s range, as Sarah wrote… However, it is worthwhile to better understand what is going on, to apply these rules/find h/experiment…
Let me know how it goes!
Mrs. Corricelli
im having trouble distinguishing when you are given SSA and there are two possible solutions and when there is only one. For instance number 11 in the ch. 6 review youare given SSA and there is only one solution, correct? why is that? when are there two?
also, mrs. corricelli, will i be able to receive extra crdit for that tangent problem on chapter 4 or 5′s test or quiz (cant remember) i got 0 points on? im assuming we’ll have something like number 17 and number 19 in the ch 6 review on tomorows quiz…..?