My Math Forum Angles and sin

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 April 20th, 2009, 01:28 PM #1 Newbie   Joined: Jan 2009 Posts: 10 Thanks: 0 Angles and sin Can someone check my work? I don't know if I did the forumlas right or not and need to know where I messed up. 1. From a point A on the ground, the angle of elevation to the top of a tall building is 24.1. From B, which is 600ft closer to the building the angle of elevation is measured to be 30.2. Estimate the height of the building. I got 1159.658507 I did this got A=24.1 B=149.8, C=6.1 then found b side by doing sin(6.1)/600 = sin149.8/b so b= 2840 then I got the new angles for the bigger triangle and got A= 24.1 B=90, C=65.9 then I did Sin90/2840=sin24.1/a which is (2840sin(24.1))/sin(90) = 1159.658509 2. Lookouts in two fire towers, A and B located 10km apart sight a forest fire. Electronic equipment allows them to determine that the fire is at an angle of 71 from tower a and 100 from tower b. Which tower is closer to the fire and how far away is it? I eventually got it to sin(71)/a=sin(9)/10 = 60.44 3. Solve sin2x+cosx=0 in the interval [0,2pie) 2sinx+cos cos(2sin+1)=0 cos=0 sin=-1/2 so, Pie/2,3pie/2,4pie/3,11pie/6 4. If z= (1/2)+(square root 3/2)i ,then z^16= I got (1/65536)+(6561/65536)i I just put the power over everything.
 April 20th, 2009, 02:07 PM #2 Senior Member   Joined: Feb 2009 From: Adelaide, Australia Posts: 1,519 Thanks: 3 Re: Angles and sin 1. Right working. More accurately 1159.743 2. The question is ambiguous without a diagram. Using the sine law and making some guesses about what is meant, I find B is nearer, 10sin(80)/sin(29) km 3. Not 4pi/3 but 7pi/6, rest is ok 4. This complex number is on the unit circle. It equals $e^{i\pi/3}$ (can you figure out why?) So taking it to the power of 16 you will get $e^{i16\pi/3}$, which is still on the unit circle... use geometry to convert to the form x+yi.
 April 20th, 2009, 06:07 PM #3 Senior Member   Joined: Jul 2008 Posts: 895 Thanks: 0 Re: Angles and sin #1. This is simpler if you let the distance from the closer to the building be "d", and the height "h". Then use cotangents for the two right triangles, then eliminate "d" and solve for "h". #2. Aswoods is right about the ambiguity. A diagram would help. WE coudl assume that the angles are both within the triangle, but that may not be so. #3. Quite a jump from applied trig to complex numbers!! Which right triangle has sides in the ratios given? If you know that, you can then look up DeMoivre's Theorem.

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