The length of arc ACB is 57.2.
Semicircle
If angle ACB is the right angle then ab is the hypotenuse. Then, (ab)2 = 62 + 92 = 36 + 81 = 117 ab = √117 = 10.8 (3 sf) If angle BAC is the right angle then ab is one leg of a right angled triangle with bc the hypotenuse. 92 = 62 + (ab)2 : (ab)2 = 92 - 62 = 81 - 36 = 45 ab = √45 = 6.71 (3 sf)
The picture that you see printed next to that question on the assignment sheet is necessary in order to answer it. Without the picture, I have no idea how angle-1, arc-ab, and arc-ag are connected .
Place the point if the compass on point B and draw an arc across AB.
The length of arc ACB is 57.2.
Arc AB represents 40/240 = 1/6 of the circumference of the circle. As the angle at the centre subtended by the whole circle is 360° then ∠A0B (if the center is O) measures 1/6 x 360 = 60°. Since a central angle has the same number of degrees as the arc it intercepts, the arc ACB (note we can call the arc AB as arc ACB) measures 60°.
The equation for finding the length of an arc is S=rθ,where S is the arc length, r is the radius, and θ is the angle in radians.Assuming you mean AOB=240 is the angle of the arc you are measuring in degrees:θ=(pi*240)/180=4.1888radTherefore the arc length is 4cm*4.1888rad=16pi/3=16.76cm
length work
Ab+ac+5mp=acb
circumference of the circle = 2*pi*10 = 20pi units of measurement length of arc = (120/360)*20pi = 20.944 units (rounded to 3 decimal places)
30
Semicircle
If angle ACB is the right angle then ab is the hypotenuse. Then, (ab)2 = 62 + 92 = 36 + 81 = 117 ab = √117 = 10.8 (3 sf) If angle BAC is the right angle then ab is one leg of a right angled triangle with bc the hypotenuse. 92 = 62 + (ab)2 : (ab)2 = 92 - 62 = 81 - 36 = 45 ab = √45 = 6.71 (3 sf)
The picture that you see printed next to that question on the assignment sheet is necessary in order to answer it. Without the picture, I have no idea how angle-1, arc-ab, and arc-ag are connected .
Place the point if the compass on point B and draw an arc across AB.
Using only a compass you cannot do it - you will need a straight edge as well.Draw a straight line AB.From A, using the length AB draw an arc above AB.From B, using the length AB draw an arc to intersect the first arc at C.Join CA.Then angle CAB is 60%From A, draw an arc to cut the two arm of the angle, AB and AC at X and Y respectively.From X, draw an arc between the two arms of angle CAB.Using the same length, repeat from Y so as to cut the previous arc at D.Join DA.DA bisects angle CAB and so angle DAB = 30 degrees.Now bisect angle DAB to give angle EAB = 15 degrees.At A, draw a perpendicular (facing downwards). If F is any point on the perpendicular, the angle EAF = 15 + 90 = 105 degrees.