Recent Homework Questions About Trigonometry
Trigonometry Questions 1.) Find the exaqct value of tan(11π/12) 2.) A linear trig equation involving cosx has a solution of π/6. Name three other possible solutions 3) Solve 10cosx=-7 where 0≤x≤2π
Wednesday, December 4, 2013 at 2:13pm
make a diagram showing a sideview. remember the angle of depression from the helicopter view is equal to the angle of elevation from groundview. So I have a right-angled triangle, with a height of 105 (opposite side), the adjacent side of x and an angle of 30.5 take it from there
Wednesday, December 4, 2013 at 8:19am
Math - Trig
Looks like you need to review the basic trig functions and draw useful diagrams. #1 h/15 = tan 46.48° #2 h/17.2 = tan 73.5°
Wednesday, December 4, 2013 at 12:11am
You have received many replies to your numerous trig questions, and have not responded to a single one of them. How do we know that you are even reading the responses, and if so, if they are of any help to you. What is your main difficulty with these questions? What are you ...
Thursday, November 28, 2013 at 10:53pm
David, really now ! After getting help for so many trig question, don't you think it is time that you try some of these yourself? Especially the rather simple ones, like your last two? Let us know what you get and we'll check them
Thursday, November 28, 2013 at 9:08pm
if cot 2θ = 5/12, the 2θ is in QI or QIII. But, since 2θ<π, we must be in QI. so, sin2θ = 12/13, cos2θ = 12/13 Now use the half-angle formulas to find sinθ, cosθ, tanθ.
Wednesday, November 27, 2013 at 11:57pm
The tree top is 100*tan(52) feet higher than the person's eyes. To get the height h, above ground, h=100tan(52)+5 feet = 133 feet, approximately
Wednesday, November 27, 2013 at 4:31pm
Math - Trig
RS/sin T = RT/sin S now just plug in your numbers.
Wednesday, November 27, 2013 at 2:01am
Math - Trig
since the altitude divides the base in half, and the vertex angle in half, h/92.1 = cot 21°55'
Wednesday, November 27, 2013 at 1:58am
Do you not have a calculator? if cotØ = .791 then tanØ = 1/.791 = appr 1.264 Ø = 51.656° secØ = 1/cosØ = 1.612 2. sin^2 Ø + cos^2 Ø = 1 , one of the major trig identities, which you MUST know.
Tuesday, November 26, 2013 at 10:51pm
Trig-Geometry - Law of sines and cosines
Hello everyone, I've been struggling on this problem for quite some time. It would be appreciated if you could help. Thanks. (The website is the diagram, it is a screenshot) In the diagram below, triangle ABC has been reflected over its median AM to produce triangle AB'...
Sunday, November 24, 2013 at 11:01pm
the ratio of height to shadow length is the same, so h/3.5 = 8/5 I imagine you can solve that for h.
Sunday, November 24, 2013 at 6:06pm
if you don't understand it at all, you seriously need to review the most basic trig identities: sin^2 + cos^2 = 1 tan = sin/cos cot = 1/tan With that in mind, and suppressing all the x's for ease of reading, (a) sin*tan = sin/cot = sin*cot/cot^2 = sin(cos/sin) / cot^2...
Sunday, November 24, 2013 at 4:10pm