All others can be derived from these and a little calculus: sin2x+cos2x=1 sec2x-tan2x=1 sin(a+b)=sin(a)cos(b)+sin(b)sin(a) cos(a+b)=cos(a)cos(b)-sin(a)sin(b) eix=cos(x)+i*sin(x)
Yes, but you would need to know a degree measure too. [Sin(A)/a] = [Sin(B)/b] = [Sin(C)/c] [a/Sin(A)] = [b/Sin(B)] = [c/Sin(C)]
Consider any triangle ABC, and let AD be the altitude from A on to BC. Then sin(B) = AD/AB so that AD = AB*sin(B) and sin(C) = AD/AC so that AD = AC*sin(C) Therefore AB*sin(B) = AC*sin(C) or c*sin(B) = b*sin(C) where the lower case letter represents the side opposite the angle with the upper case name. Divide both sides by bc to give sin(B)/b = sin(C)/c. Similarly, using the altitude from B you can show that sin(A)/a = sin(C)/c. Combining with the previous result, sin(A)/a = sin(B)/b = sin(C)/c.
First, note that sin(a+b)=sin(a)cos(b)+sin(b)cos(a)[For a proof, see: www.mathsroom.co.uk/downloads/Compound_Angle_Proof.pptFor the case of b=a, we have:sin (a+a)=sin(a)cos(a)+sin(a)cos(a)sin (2a)=2*sin(a)cos(a)
You need to use the sine rule. If the three angles are A, B and C and the sides opposite them are named a, b and c then, by the sine rule, a/sin(A) = b/sin(b) = c/sin(C) Therefore b = a*sin(B)/sin(A) = a*y where y = sin(B)/sin(A) can be calculated and c = a*sin(C)/sin(A) = a*z where z = sin(C)/sin(A) can be calculated. then perimeter = p = a + b + c = a + ay + az = a*(1 + y + z) therefore a = p/(1 + y + z) or a = p/[1 + sin(B)/sin(A) + sin(C)/sin(A)]. Everything on the right hand side is known and so a can be calculated. Once that has been done, b = a*y and c = a*z.
Thanks to the pre-existing addition and subtraction theorums, we can establish the identity:sin(a+b) = sin(a)cos(b)+sin(a)cos(b)Then, solving this, we getsin(a+b) = 2(sin(a)cos(b))sin(a)cos(b) = sin(a+b)/2a=b, sosin(a)cos(a) = sin(a+a)/2sin(a)cos(a) = sin(2a)/2Therefore, the answer is sin(2a)/2.
The sum of tthe angles of a triangle is 180° which means the third angle is 180° - (57° + 71°) = 52° The sine rule gives: a/sin A = b/sin B = c / sin C where side a is opposites angle A, etc. The sine rule can be used to find the lengths of the other two sides when the angles are all known and one side length is known. Let angle A = 57°, then side a = 14.5 in. Let angle B = 71° and angle C = 52° Using the sine rule: a/sin A = b/ sin B → b = a × sin B/sin A Similarly, c = a × sin C/sin A → The perimeter = a + b + c = a + a × sin B/sin A + a × sin C/sin A = a(1 + sin B/sin A + sin C/sin A) = 14.5 in × (1 + sin 71° / sin 57° + sin 52° / sin 57°) ≈ 44.47 in ≈ 44.5 in
By the sine rule, sin(C)/c = sin(B)/b so sin(C) = 25/15*sin(32d15m) = 0.8894 so C = 62.8 deg or 117.2 deg. Therefore, A = 180 - (B+C) = 85.0 deg or 30.5 deg and then, using the sine rule again, a/sin(A) = b/sin(B) so a = sin(A)*b/sin(B) = 28 or a = 14.3
The Law of Sines: with triangle ABC, the angles are A, B, & C. The sides {a, b, & c} are opposite of the respective capital letter vertex.a/sin(A) = b/sin(B) = c/sin(C). You know the angles A, B, C; and two sides (say a & b).So side c = a*sin(C)/sin(A) = b*sin(C)/sin(B).You could also use the Law of Cosines: c^2 = a^2 + b^2 - 2*a*b*cos(C)
The best way to answer this question is with the angle addition formulas. Sin(a + b) = sin(a)cos(b) + cos(a)sin(b) and cos(a + b) = cos(a)cos(b) - sin(a)sin(b). If you compute this repeatedly until you get sin(3x)cos(4x) = 3sin(x) - 28sin^3(x) + 56sin^5(x) - 32sin^7(x).
Perhaps you can ask the angel to shed some divine light on the question! Suppose the base is BC, with angle B = 75 degrees angle C = 30 degrees then that angle A = 180 - (75+30) = 75 degrees. Suppose the side opposite angle A is of length a mm, the side opposite angle B is b mm and the side opposite angle C is c mm. Then by the sine rule a/sin(A) = b/(sin(B) = c/sin(C) This gives b = a*sin(B)/sin(A) and c = a*sin(C)/sin(A) Therefore, perimeter = 150 mm = a+b+c = a/sin(A) + a*sin(B)/sin(A) + a*sin(C)/sin(A) so 150 = a*{1/sin(A) + sin(B)/sin(A) + sin(C)/sin(A)} or 150 = a{x} where every term for x is known. This equation can be solved for a. So draw the base of length a. At one end, draw an angle of 75 degrees, at the other one of 30 degrees and that is it!
b/c