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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
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)]
How do you get the sides of the triangle when 3 angles are given and its perimeter?
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.
What is the length of side c in Triangle ABC to the nearest whole number if A equals 42 degrees and B equals 87 degrees and a equals 24?
28 The Law of Sines: a/sin A = b/sin B = c/sin C 24/sin 42˚ = c/sin (180˚ - 42˚ - 87˚) since there are 180˚ in a triangle. 24/sin 42˚ = c/sin 51˚ c = 24(sin 51˚)/sin 42˚ ≈ 28
Explain what the Law of sines becomes when one of the angles is a right angle?
The Law of sines: a/sin A = b/sin B = c/sin CIf the angle C in the triangle ABC is 90 degrees, then the triangle ABC is a right triangle, where c is the measure of the hypotenuse, a is the measure of the leg opposite the angle A, and b is the measure of the leg opposite the angle B.Let us observe what happens when sin C = sin 90 degrees = 1.c/sin C = a/sin A cross multiply;c sin A = a sin C divide by c both sides;(c sin A)/c = (a sin C)/c simplify c on the left hand side;sin A = (a sin C)/c = [(a)(1)]/c = a/csin A = (measure of leg opposite the angle A)/(measure of hypotenuse)From the Law of Cosine we know that cos A= (b^2 + c^2 - a^2)/(2bc). If we substitute a^2 + b^2 for c^2, we have:cos A = (b^2 + (a^2+ b^2) - a^2 )/(2ab)cos A = 2b^2 /2ab simplify;cos A = b/c = (measure of leg adjacent the angle A)/(measure of hypotenuse) Therefore tan A = sin A/cos A =(a/c)/(b/c) = (a/c)(c/b) = a/b = (measure of leg opposite the angle A)/(measure of leg adjacent to angle A). And cot A = cos A/sin A = (b/c)/(a/c) = (b/c)(c/a) = b/a = (measure of leg adjacent to angle A)/(measure of leg opposite the angle A).
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!
How do you find the length of third side when 2 sides and all three angles are known?
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)
Can an equiangular triangle be scalene?
No. An equiangular triangle is always equilateral. This can be proven by the Law of Sines, which states that sin A / a = sin B / b = sin C / c, where A, B and C are angles of a triangle and a, b and c are the opposing sides of their corresponding angles. If A = B = C, then sin A = sin B = sin C. Therefore for the equation to work out, a = b = c. Therefore the eqiangular triangle is equilateral, and therefore not scalene, which requires that all sides of the triangle be of different lengths.
How do you find the length of a triangle sides using perimeter and angles?
The solution relies on using the sine rule.Suppose that the perimeter of triangle ABC is P.Then you need to divide P into 3 parts in the ratio of sin(A) : sin(B) : sin(C).Let sin(A) + sin(B) + sin(C) = X. ThenAB = P*sin(C)/XBC = P*sin(A)/XCA = P*sin(B)/X
In a triangle ABC b equals 15 cm and c equals 25 cm and also angle B equals 32'15'Find the side a and other angles?
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
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
