H Algebra 2 Trig Ch. 15

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H Algebra 2 Trig Ch. 15

4/19/2011

Law of Sines and Law of Cosines! Yes! Happy Blogging!

Archana Sathappan

4/19/2011 11:53:12 am

Can someone please help me on number 18 part b and c on pg. 868?

Johannes Grandin

4/20/2011 03:47:14 am

For those draw a triangle first to help you and remember that Wendy walked 101 meters since each pace was one meter.

Ishta

4/20/2011 12:42:04 pm

I tried doing question 18 again, but I still can't solve part b and c. Can someone help? How do I draw the triangle for the two parts. Thanks.:)

Johannes Grandin

4/21/2011 04:01:07 am

Does anyone understand numbers 1, 3, and 7 on last night's 15.2 homework?

Ishta

4/23/2011 08:16:08 am

Johannes,
I can help you with 1 and 3. For 1, label the triangle so angle A is 51. So your formula would be a2=b2+c2-2bccosA. SO when you plug it in, you have a2=4squared +5squared-2(4)(5)cos(51). Solve it and voila. The answer is 3.91 Number is 3 is similar, angle A is 138. The equation is r2=p2+q2-2pqcosR so plug it in and r2=3squared+2squared-2(2)(3)cos(138). Solve it and voila-the answer is 4.68.

Johannes Grandin

4/26/2011 03:45:44 am

Thank you.

Archana Sathappan

4/26/2011 11:50:37 am

Can someone please help me on pg. 919 # 13? For part a, I started it out, like they said by splitting it into 2 triangles and for one triangle I got an area of 1560.6 m^2, but I was unable to figure out how to find the area for the other triangle. Any help would be great, thanks!

Ishta

4/28/2011 08:38:12 am

Archana,
This question is tricky. The question stated to draw a quadrilateral that is not a rectangle, but it did not say a trapezoid (I think I spelled it wrong). So draw a trapezoid where the bases are both 50. That's what I did. I think this is not the best explanation but I can't think of any other way to solve the problem.

Rohan Roy

4/28/2011 08:57:29 am

So whenever you are trying to find a angles, for either right triangles, law of sines or cosines, you always use the inverse button right?

Archana Sathappan

4/28/2011 10:04:35 am

Rohan,
Yes, when finding angles use the inverse.

Nikita Hariharan

5/7/2011 04:31:27 am

Ishta:
You can draw your triangles in a similar way to the problem right before it. The way they did the cross intersection of th estreets for #17. Does that make any sense?

Nikita Hariharan

5/7/2011 12:02:35 pm

Hey can anybody explain ambiguous cases to me? I understand everything except the ambiguous case for the test.

Rohan Roy

5/8/2011 04:08:08 am

So for the ambiguous case, if you get a degree under 180 then there is only one possibility, and if its over 180 theres no solution, and if it perfectly adds up to 180 then theres 2 solutions. is that right?

Ishta

5/8/2011 09:09:32 am

I agree with you Rohan. To find the ambiguous case, subtract the ambiguous angle from 180. Then, use the given info and see if it is over/under 180 to trell. Does that help, Nikita?

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H Algebra 2 Trig Ch. 15

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