Find the angle subtended by the chord and the length of the chord.
203052
The angle is 36 degrees (10%). The chord length is 30.901699 cm = 2*50*cos(90-36/2) cm.
339660
There's not enough information to answer the question. We need to know at least one angle as well as the length of two sides.
358979
the angles are just unknown if i get you correctly.\nthere are sin, cos, and tan proportions you can work out to find the rest. i think you only need three of the measurements to work out the rest of the triangle, like 2 sides and an angle or 2 angles and a side.  not positive.
217054
Consider a triangle with sides a, b, c and angle A oppose to the side a.  Side a is the diagonal that you are going to find in each case.\n\nBy knowing 2 sides (b, c) and the angle (A) in between, then use Cosine rule:\n\na^2 = b^2 + c^2 - 2bc cos A\n\na^2 = 60^2 + 80^2 - 2 * 60 * 80 cos(56)\n\na = 68.06 ft  (shorter diagonal)\n\nConsider when A = 180 - 56 = 124, use Cosine rule again:\n\na^2 = 60^2 + 80^2 - 2 * 60 * 80 cos(124)\n\na = 123.97 ft (longer diagonal)
283153
Use sin formula first:\n1. find the other angle (180-(15-30), remember sum of 3 angle in triangle is 180 degree)\n2. a/sinA=b/sinB (need only find one other side)\n3. Area = 1/2ab.sinC\n\nAnswer:(sqrt(3)+1)/4 (not in the option!)
284681
I wish I remembered terms for this stuff.\nFind the length of the base, divide that by two.\nUse this number to find the length of a line drawn from the point of the triangle to the base. Congrats, that number is the height.\nTriangle ABC.\nLength from point A to C is 3.\nFrom A to B is 4.\nSo B to C is 5.\nTake the base, we'll say line AC, and divide that by 2, so you get 1.5.\nUse that number, now, as the length from A to D.\nSo, 1.5 squared plus 4 squared is 18.25. Square root of 18.25 is 4.27, so your height is 4.27.\nPythagorean Theorem.\nA^2 + B^2 = C^2
421856
I'm not sure how far I'll get with this one, but I'll give it a try.\n\nLet the center of the circle be at (p,0), so the equation of the circle is (x-p)^2 + y^2 = a^2, with a>p.  The chord is y = mx, and we want to find where the chord and circle intersect.\n\n(x-p)^2 + m^2 x^2 = a^2\nx^2 - 2px + p^2 + m^2 x^2 = a^2\n(m^2 + 1) x^2 - 2px = a^2 - p^2\n\nSolve by completing the square (details omitted) to get\n\nx = {p +/- sqrt[m^2 (a^2 - p^2) + a^2]} / (m^2 + 1)\n\nThe center of the chord (hence center of the circle with chord as diameter) is at x = p / (m^2 + 1), with y = mx = mp / (m^2 + 1).  The radius is sqrt[m^2 (a^2 - p^2) + a^2] / (m^2 + 1), so the equation of that circle is\n\n[x - p / (m^2 + 1)]^2 + [y - mp / (m^2 + 1)]^2\n  = [m^2 (a^2 - p^2) + a^2] / (m^2 + 1)^2\n\nIf q = m^2 + 1, then by clearing fractions, we get\n\n(qx - p)^2 + (qy - mp)^2 = m^2 (a^2 - p^2) + a^2\n\nI'm not going to check my work, but those are the answers I ended up with.
209620
Use the double angle identities.
634140
Brian gave you the answer, but he didn't explain it thoroughly.\n\nPlace your vector at the origin of an x and y plane.  The x and y axes are perpendicular, so you can use the Pythagorean formula to get the magnitude which is the length of the hypotenuse of the right triangle formed by the points (0,0), (10,0), and (0,8). sqrt(10^2+8^2)\n\nNow you need direction which is the angle between the x axis and the vector. Since you have solved for all three lengths of the triangles sides, you can use any inverse trig function to find the angle.  Brian chose tangent.\n\nGood work, Brian!
547300
sorry, i don't have a calculator w/ me right now but i'l try\n1. use the law of cosines\na squared = b squared + c squared -2bc*cosA\n2. in a 30-60-90 triangle the sides are x- the side across from the 30; x*square root 3- the side across from the 60; 2x-the side across from the 90\n3. i would try the law of sines\na/sinA = b/sinB = c/sinC\nremember that you know two angles, the 90 and the 61\nso the 61 is the degree the ladder and the ground forms.  set up the problem like this: x/sin61 = 18/sin90\nit should work\n\nto find secant, type into the calculator 1/cos(36) and check what mode you're in (degree or radians)\ni'm pretty sure cot60 = 1/the square root of 3\nsec30=2/the square root of three
376123
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.
222717
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.
513971
Use the fact that the sum of the angles of a triangle add up to 180 degrees. Let A, B, C be the angles of the triangle. The m<A + m<B + m<C =180. If you know m<A and m<B, you can find m<C.\n\nWhen I write m<A, I mean the measure of angle A, not m is less than A.
320537
There is no work to show, rather theorems to justify the solutions.\n\nInscribed angles of a circle that subtend a diameter are 90 degrees. This is a corollary of a theorem which states that inscribed angles are 1/2 the measure of a sector angle which subtends the same arc. In the diameter case the sector angle is 180 degrees, thus the inscribed angles are 90.\n\nThe quadrilateral that contains angle Z has it's vertices on the circumference and is cyclic. A cyclic quadrilateral has the property that opposite angles are supplimentary (add  up to 180) thus angle Z = 180 - 148 = 32 degrees.\n\nI am not sure how you would show work toher than proving the theorems.\n\nHope it helps
524837
A short element of the curve is dL=sqrt(dr^2+(r*dt)^2), r*dt elementary length of the arc with radius r and polar angle t (your theta), dr elementary increase of r for dt.\nThus dL=sqrt(r’^2+r^2)*dt.\nIf  r=a*(1-cost), then dL=a*sqrt((sint)^2+(1-cost)^2)=a*sqrt(2)*sqrt(1-cost)*dt. \nFortunately 1-cost=2*(sin(t/2))^2 or dL=2a*sin(t/2)*dt, L=-4a*cos(t/2) for t=0 until 2pi, thus L=-4*a*(-1-1)=8*a – Oh! check it poor kid!\nIf r=(sin(t/2))^2=0.5*(1-cost), then a=0.5 for t=0 until pi. It must be done likewise. Thus L=-4*0.5*(0-1)=2
561616
Draw a picture of the two congruent sections, which are connected at B. The exact angle between them doesn't matter so long as it's not greater than 90 degrees. You will see that if AD=CD, then angles A and C are equal. So what happens as angle A gets larger and angle C gets smaller? You should be able to see the answer for yourself.
497877
So give the first one x+6, the second one x, and the third one 3(x+x+6)-4\nnow the second step, x+(x+6)+(6x+14)=180\nnow, 8x+20=180\nthen, 8x=180-20\n         8x=160\n         x=160/8\n         x=20.\ntherefore the first angle =26, second, 20 and the third 134