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How do you derived the sine law?
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.
Can you use the law of sines if 3 sides are given?
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)]
What is the perimeter of a triangle when an angle of 57 degrees is opposite to a side of 14.5 inches and has another angle of 71 degrees?
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
How do you construct a triangle with perimeter 150 mm and a base angle 75 degrees and 30 degrees?
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!
What are the ratios of the length of the longer leg of a 30-60-90 triangle to the length of its hypotenuse?
The longer leg is opposite the 60 deg angle. Suppose A = 60 deg, C = 90 deg and a and c are the corresponding sides. Then, by the sine rule a/c = sin(A)/sin(C) a/c = sin(60)/sin(90) = sqrt(3)/2
