You are at the top of a buidling 418 ft tall. As you look down at anearby smaller buidling, you observe that the angle of depression of the base is 77 degrees and the angle of depression of the top is 46 degrees.  Find the hegiht of the smaller building and the horizontal distance between the two buildings.
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Trigonometry is hard to do in you head, so start out by making a drawing. You have two buildings A (tall) and B (short) - Draw them far apart, not like this:\n\nA         B\n|         \n|         \n|    \n|         |\n|         |\n|         |\n|         |\n\nNow draw one line from the top of A to the top of B - this is the angle of depression of the top and it is 46 degrees. That is, if you drew a straight, horizontal line across the top of building A, the angle between that line and the line joining the top of A to the top of B is 46 degress.\n\nSimilarly, draw a line from the top of A to the *bottom* of B - this is the angle of depression of the bottom and it is 77 degrees. That is, the angle between the horizontal line across the top of A, and the line between top of A and bottom of B is 77 degrees.\n\nNow can you find out the angle between the line from A-top to B-top, and the line from A-top to B-bottom? (It's very important for you to draw this!) Let's put this value aside for a minute. \n\nLooking at your drawing, can you see that you know the value for *part* of the angle between the horizontal line across A's top and length of building A? What is that part you know? Now, see how the line across A's top and length of A make a right angle? So how many degrees in a right angle? So if you know the total of that angle, and part of that angle, how do you get the other part? **Notice that this part we're trying to get is actually the angle between the line A-top to *B*-bottom, and the line A-top to *A*-bottom. \n\nYou should also notice that now we have two triangles:\nTriangle 1: A-top to B-bottom to A-bottom. Solve this triangle for line A-bottom to B-bottom for distance between the two buildings.\nTriangle 2: A-top to B-top to B-bottom. Solve this triangle for line B-top to B-bottom for height of building. \n\nDon't forget, we have two important angles that will help solve these two triangles!
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Since the height is the only thing considered you can disregard the initial 32ft/sec (this is the initial x direction velocity).  When the height is 12ft you set the equation equal to 12 and use the quadratic formula to solve for t {[at^2+bt+c=0; a=-16, b=32, c=-12] stick these values into    (-b+/-sqrt(b^2-4ac))/2a} You'll find that t=1/2,3/2.  He reaches 12 feet going up at .5 secs and again coming down at 1.5 secs.  Do the same for 14ft and you should get two answers (0.646s,1.35s) He is above 14ft between these times so you subtract the larger time from the smaller and you get about 0.707 secs.\nIf you have a problem understanding my notation check out the quadratic Equation at http://mathworld.wolfram.com/QuadraticEquation.html
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this is a simple dervitive problem....we recall that given a position the first derivitive would give a velocity at any time t, and a second derivitive would give an accelaration at any time t\n\nso we need to take the derivitive of h(t)\n\nso h(t) = -6.5t^2 +40t+6\nh'(t)= -13 t +40\n\nnow you find when h'(t) is zero...this will tell you when the slope of a tangent line of position is zero...in otherwords...a max or min\n\nso you find h'(t)=0=-13t+40\nso t= 40/13\n\nnow we need the second derivitive to  apply the second derivitive test (recall this tells us that if the h''(t) is negative at a specific t value we have a max, and if its positive at that t value its a min)\n\nso h''(t)=-13, so for all values of t h''(t) is negative so at t=40 h'(t) is zero and h''(t) is negative so there is a max height at t= 40/13\n\nnow to find the height you plug 40/13 into h(t)\n\nso h(40/13) i will leave to you cause this is teh easy part\n\nhope this helps\n\nmatttlocke
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v = v0 - at\n0 ft = 288 ft/sec - 32ft/sec² t\nt =  288 ft/sec / 32ft/sec²\nt= 9 sec\n\ns = 1/2at²\ns = 1/2 (32ft/sec²) (9 sec)²\ns = 1,296 ft\n\nIt's the same time up and down, so 18sec until it hits the ground.
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..... ^\n... / .. &#92;\n.. / A1 &#92;\n../-------&#92;\n./ .. A2.. &#92;\n/______&#92;\n\nIs this what you mean (where A1/A2 = 1/φ; and φ = golden ratio = ½(1 + √5))?\n\nIt is my understanding.\n\nSo the two triangles are similar \n\nArea = ½ base * height\n\nA1 / A2 = 1/φ\nThus A1/ (A1 + A2) = 1/(1 + φ)\nie A1 /A = 1/(1 + φ)\n\nThus corresponding sides are in the ratio 1/√(1 + φ) \nNow √(1 + φ) = φ \n(since φ - 1 = 1/φ then φ² - φ = 1 so φ² = φ + 1)\nSo the top base = 1/φ * 8 inches ≈ 4.94 inches\nand the top side walls = 1/2 * 8/φ * tan 52°\n≈ 3.16 inches\nThus the bottom side walls are 8 - 1/2 * 8/φ * tan 52°≈ 4.84 inches
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a) You have to draw gravity as a straight line down. Then you have to realize that gravity does no work if the block was on a horizontal surface. Work done by gravity is alway M*G*H.  The mass is 4.8 kg and the gravity is 9.81 m/s^2 .  The height will be sin (30) = H/2.  the sin (30 degrees) is 0.5 so .5=H/2.  the height will be .5/2 = .25 m.\n\nsoo you will have 4.8*9.81*0.25\n\nb) the work done by the frictional force will always be the frictional force times the distance.  The frictional force is going to be the normal force times Uk.  The normal force has components to it, so its going to be sin(30)*g*m.  The you need to multiply that by the friction coefficient and then by the distance traveled.  You will have 2*0.436*sin(30)*4.8*9.8\n\nc.  The work done by the normal force is just going to be cos(30) (i think) times 2 ( the distance traveled) times 9.81 (gravity) times 4.8 (the mass)
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1. Find x to the nearest length\nWhat would x be if it is a right triangle with the cos angle at 32º,sin angle at 58º and a hypotenuse length at 10\n\ncos 32 = x/10\n\nSo x = 8.48\n\n2.Find the missing side lengths for x and y.\nthe triangle is a 30-60-90 and it has a hypotenuse of 8. the sin leg is x and the cos leg is y\n\nsin30=x/8\nx = 4\n\ncos30 = y/8\ny = 4√3\n\n3. a ladder 18 feet long makes an angle of 61º with the ground as it leans against a barn. How far up the side of the barn does the ladder reach?\n\nThe ladder leaning against the barn makes a right-angled triangle with a hypotenuse of 18.\n\nLet x be the height of the triangle. Then:\n\nsin61 = x/18\nx = 15.74 feet\n\n4. also, to make sure i did things correctly, is the Sec36º -.128? and is the cot60º the squareroot of 3 /3?\n\nNo, that is incorrect.\n\nsec 36 = 1/cos36 = 1.236\n\ncot60 = cos60/sin60 = (1/2)/(√3/2) = 1/√3\n\n5. Is sec 30º =2? \n\nsec30 = 1/cos30 = 1/sin30 = 2
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actually its not\n the answer would be (sin(30degrees)*200)+5\nsin(30degrees) is .5\n.5*200 = 100 +5 = 105 feet in elevation\n\ndraw a picture and label everything as much as possible, you eralize that you have the length of a hypotenuse, so you take teh sin of the angle then multiply by the length of the hypotenuse to get the length of the opposite side, that gives you teh height from eye level, you now must add the level you are looking from (5ft)\n\nhope this helps!!
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1. So, you have two sides and included angle.  Find the 3rd side and either use Pythagorean theorem to calculate the altitude or use the tangent of 1/2 the angle and 1/2 the base.\n\n2a.  You have two sides and the included angle\n2b.  You have two angles and the included side
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height = 5sqrt(2)\nbase = 12\nArea = 5sqrt(2)*12\nArea = 60sqrt(2)\nNote that the height is neither 10 nor 12.\nIf you consider 12 to be the base, then the slant height is 10.\nArea equals base times height.\nHeight and slant height are not the same thing.
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I think you actually need to use Tangent for this problem. \n\nThe wheelbase of the truck is 1.8m. Assuming it is perfectly balanced, the center of gravity would be directly over the center of that; 0.9m from one wheel. That's the base of the triangle.\n\nThe vertical side rises up to the center of gravity; which is unknown, or X.\n\nA 30 degree tilt causes the truck to roll over. This means that the angle of our triangle is 90 degrees minus 30, or 60deg.\n\nNow we have, Tan60=X/0.9\n1.73(0.9)=X\nX=1.56
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a.  A = (pi)r^2 = 3.14 * 9 = 28.26 cm^2\nb.  A = l*w = 14 * 2 = 28 in^2\nc.  A = 1/2 base * height = 0.5 * 0.4 * 0.7 = 0.14 ft^2
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Forget the 6000N horizontal for for a second...\n\nYou have 5000N acting downward (negative) and 7000N acting upward (positive). The total net force acting vertically to the A/C is 7000-5000= +2000N so upward. We know that the horizontal force of 6000N is acting at a 90 degree angle to the upward force so the resulting forces make a right triangle. Use Pathagorean theorem.\n\n2000^2 + 6000^2 = x^2\n\nx is your MAGNITUDE.\n\nNow find the angle:\n\nUse trigonometry: \n\nTan(Angle) = opposite/adjacent\nwe know the opposite is 2000 and the adjacent is 6000 so,\ninverse Tan(2000/6000) = 18.4 degrees, upward and forward.
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I can only answer 1 and 3 for you, cause i dont know the others, but here it goes:\n\n             1.) Think of this problem in terms of a right triangle. Draw a horizontal line. This line is the tidal force of 3* (10^3) N. Then draw a line on the end of the line perpendicular to the horizontal one. This line is your wind force of 6* (10^3) N. Connect the two ends of the lines with another line, to make a complete right angle triangle. This is where it gets a little tricky. The angle formed when the slanted line and the horizontal line meet is the angle you are trying to find. To find it, use the tan function, which is defined as opposite side's value (6*10^3 N) over the adjacent side's value (3*10^3 N). Lets use x as a variable for the unknown angle. So...\n\ntanx=(6*10^3) / (3*10^3)\nUse inverse of tan now (on calculators, you usually have to press shift or 2nd, then the tan key.) So...\n\nInverse tan ( (6*10^3) / (3*10^3) ) = x\n\ndo that in your calculator, and the value you get is the angle.\n\nFor the net force just subtract (3* 10^3) from (6*10^3) and the result will be your net force.\n\n3.) I dont know if u know this or not, but there is a formula that is part of Newton's second law. It is F=m*a.\n"F" being the Force, while "m" is the mass, and "a" is the acceleration. Just plug in the values you are given so that it comes out....\n\n390=270*a\n\n390/270=a ("a" is in meters per second squared)\n\nHope that helps
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45 degrees is the best angle in theory.\n\nIn practice you may have a adjust a little to compensate for air resistance and wind factor.
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assuming your 30 degree depression is from the normal line of sight, then about 86.6 m (50 * sqrt(3))
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a)A=41°20’ b=17.2cm\ncos 41°20'=17.2/h\nh=17.2/cos 41 1/3°=22.9cm\n\nb) 1) A=48°40’ a=21.0cm\nsin 48 2/3°=21/h\nh=21/sin 48 2/3°=28.0cm\n\nsolve TRIANGLE ABC if \nA=50° B=81° a=95cm\n\nC=180-50-81=49°\nlaw of sines\n95/sin 50=b/sin 81=c/sin 49\nb=95 sin 81/sin 50=122.5cm\nc=95 sin 49/sin 50=93.6cm