In triangle ABC, sin n=0.891. Which is closest to csc n?
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The answer is B) 1.122\n\nFor the different trigo. functions:\n\nLet:\na = side opposite angle n\nb = side adjacent angle n\nc = hypotenuse of the right triangle\n\nTherefore:\n\ncos n = adj./hyp. = b/c\nsin n = opp./hyp. = a/c\ntan n = opp./adj. = a/b\nsec n = hyp./adj. = c/b\ncsc n = hyp./opp. = c/a\ncot n = adj./opp. = b/a\n\nYou will notice the following:\n\n1) the secant is the reciprocal of cosine.\n2) the cosecant is the reciprocal of sine.\n3) the cotangent is the reciprocal of tangent.\n\nTherefore: \n\ncsc n = 1/sin n = 1/0.891 = 1.122\n\nFor the remaining trigo. functions, solve for n first:\n\nn = arcsin 0.891 = 62.99917663˚ or 62˚59'57.04"\n\nTherefore:\n\n1) cos 62˚59'57.04" = 0.454\n2) tan 62˚59'57.04" = 1.963\n3) sec 62˚59'57.04" = 1/cos 62˚59'57.04" = 2.203\n4) cot 62˚59'57.04" = 1/tan 62˚59'57.04" = 0.510\n\nI hope that the solution was clear enough...
745771
first find the midpoint of AB which is C(9,-2).\nNow obviously the one-fourth distance will be the midpoint of AC which is ...well I let you do this. This is simple knowing the formula for the midpoint .\nMathguy did well too but you are supposed to apply, I think, midpoint formula.
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D\n\nSeriously, what class are you in? Algebra? Geometry? You chould know this!! It's the pythagorean theorm- [a2 +b2 =c2] (2's = squared).  You have a and b, the two adjacent sides of the triangle.  Therefore, it is the square root of 50, or approx. 7.07.  Thats SUPER important, make sure you remember it!
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Distance formula=sq.root of{(x2-x1)sq.+(y2-y1)sq.}\n\nDistance AB=(from formula) sq.rt. of 89\nDistance BC=(from formula) 10\nDistance AC=(from formula) sq.rt. of 89\n\nFrom above-AB=AC.\nTwo sides are equal hence ABC is an isoceles triangle.
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No calculator required.\n\nLet x = arccos(sin(3pi/7))  Then\ncosx = sin(3pi/7)\n\nNow we have the equality cosx = sin(pi/2 - x), so\nsin(pi/2 - x) = sin(3pi/7)\npi/2 - x = 3pi/7\nx = pi/14.\n\nYou can also think about it this way: arccos(a) means "the angle whose cosine is a."  So you're looking for the angle whose cosine is equal to the sine of 3pi/7.  If you graph angle 3pi/7 in the first quadrant and its sine, you'll see that the angle having the same cosine is the complementary angle, i.e. pi/2 - 3pi/7 = pi/14.
900391
csc^2(-30)+tan135+cos(-90)\n[csc(-30)]^2+tan(90+45)+cos90\n=4-1+0\n=3
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csc = 1/sin;  sin(3π/4) = √2/2, so 1/(√2/2) = 2/√2 = √2
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1/2 bh or bh over2.\nType triangle is inconsequential.\nb=base  h=height\nMeasure both, multiply, divide in half.
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Formula for Sine, Cosine and Tangent:\nTangent= opp./adj.\nSine= opp./hyp.\nCosine= adj./hyp
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8/5 the initial iteration (triangle)
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I think that what you really want here is the tangent ratio. You know the slope of the line, right (hint, it is right there in the equation)? You know that the tangent ratio is a over b (where a and b are the non-hypotenuse sides of the triangle), right? You should be able to use that to figure out the tangent of the angle made by the line. Now all you have to do is figure out the angle that has that number as a tangent. That would be the "tan -1" (inverse tangent) function on your calculator.\n\nIf you are working from trig tables rather than a calculator, find the number in the second column of the tangent table that is closest to your tangent number then look in the first column for the angle.\n\nIf you are required to use sin and cosine remember your Pythagorean theorem and use it to find the hypotenuse.\n\nASIDE TO Drakestone: Your solution only finds the sine of the angle, not the angle itself. You need to apply "sin -1" to your answer to get the final answer.\n\n\nASIDE TO eternal...: You are wrong too. graph y=5x+3 and you will see that it is way too steep to be an 11 degree angle. You mixed up your opposite and adjacent values in calculating your tangent.
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You're looking for the angle that has a sine of 1/2.\nDrawing a triangle representing this is very helpful.  You will see that it is a 30-60-90 triangle.  \nIf you can find it in degrees, then you can convert it to radians by multiplying it by pi/180.\nsine(30) = 1/2\n30 degrees = 30(pi/180) = pi/6
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sine 180=0\ncos 180= 1\ntan 180= 0\nsec 180= 1
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Any answer with B>90 is acceptable satisfying:\nA=2B-180\nand C=180-B-A\n\nfor example: (A,B,C)=(20,60,100) or (40,30,110)\n\nNote that ABC and DBC are similiar.
472052
ok so i assume you know the basic rules of trigonomerty for a triangle.\ni.e sinR = opposite/hypotenuse and cosR= adjacent/hypotenuse.\n\ndraw the triangle on a piece of paper.\nYou will see that sin R = ST/SR\n\n(where ST is the edge running from S to T and similarly SR is the edge running from Sto R)\n\nalso cos R= RT/SR\n\nso sinR^2 + cosR^2 = (ST)^2/(SR^2) + (RT)^2/(SR^2)\n\n= [(ST)^2 + (RT)^2]/(SR^2)\n\nNow we use pythagoras' theorem. I assume u know this? The sum of the squares of the two shorter sides of the triangle equals the square of the longer side (the hypotenuse)\n\nso (ST^2) + (RT)^2 = SR^2\n\nSO WE NOW HAVE\nsinR^2 + cosR^2 = SR^2/SR^2 = 1\n\nthere you go\n\n\nCANT E-MAIL YOU BACK WESTILE BECAUSE IT SAYS YOUR E-MAIL IS NOT VERIFIED
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There are six ratios of the three sides of a right triangle. These ratios have been identified by name throughout the centuries. In the USA they are known as sine, cosine, tangent, cotangent, secant, and cosecant. Your calculator has buttons for three of these: [SIN], [COS], and [TAN]. Remember them with the non-word SohCahToa. \n\nThe tangent ratio of an angle is simply the ratio of the opposite side to the adjacent side of a right triangle. \n\nIf you know the angle, type \n\n[TAN] > angle > [ENTER].\n \nThe answer will be the tangent (ratio of the opposite side to the adjacent side).\n\nIf you know the opposite and adjacent sides and need to find the angle, type \n\n[2nd] > [TAN] > opposite > [/] > adjacent > [ENTER]. \n\nThe answer given will be the desired angle.\n\nCaution: \n(1) Different calculators require these buttons to be pushed in a different order.\n(2) If your angle is in degrees, then make sure your calculator's mode is in degrees.\n(3) Don't get the numbers backwards or you'll find the cotangent of the angle and not the tangent.
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csc(x) is the same as 1/sin(x) and sec(x) is the same as 1/cos(x) so the left side of the equation can be simplified to \n\nsin(x)sin(x) + cos(x)cos(x) \nor sin^2(x) + cos^2(x)    {"sin squared x plus cos squared x"}\n\nUsing the same steps, the right side of the equation becomes sin(x)/sin(x) or 1.\n\nSo the identity has been reduced to  sin^2(x) + cos^2(x) = 1 which is another common trigonometry identity!