5
Use the cosine rule: a2 = b2+c2-2bc*cos A BC2 = AB2+AC2-2*AB*AC*cos A BC2 = 32+52-2*3*5*cos 57 BC2 = 17.66082895 BC = 4.20247887 cm in length
That depends on the value of CD and the perimeter of the quadrilateral out lined in the question
The probability of ac and bc is 1/5.
40?
5
AC=5 AB=8 A=1 B=8 C=5 BC=40
If AC equals 6 and BD equals 4, then AB equals 5.
9_or_yes">9 or yesA+ = 12
Use the cosine rule: a2 = b2+c2-2bc*cos A BC2 = AB2+AC2-2*AB*AC*cos A BC2 = 32+52-2*3*5*cos 57 BC2 = 17.66082895 BC = 4.20247887 cm in length
If these are sides of a triangle then AC can have any value in the interval (3, 13).
If CB is the hypotenuse, then AB measures, √ (62 - 52) = √ 11 = 3.3166 (4dp) If AB is the hypotenuse then it measures, √ (62 + 52) = √ 61 = 7.8102 (4dp)
truee buddyy (;
The length is sqrt(61) units.
Since AB and AC are tangent to the circle O, it seems that they both are drawn from the same outside point A. As tangents to a circle from an outside point are congruent, AB ≅ BC. Also, a tangent is perpendicular to radius drawn to point of contact. So that OB and OC are congruent radii. Therefore, the perimeter of the quadrilateral ABOC equals to P = 2(12 cm) + 2(5 cm) = 34 cm.
That depends on the value of CD and the perimeter of the quadrilateral out lined in the question
The probability of ac and bc is 1/5.