If a cone-shaped water cup holds 23 cubic inches and has a radius of 1 inch, what is the height of the cup?
1 answer:
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Answer:
Angle ACD = 38°
Step-by-step explanation:
The full, correct question is presented the attached image to this solution.
Given
Point O is the centre if the circle
Points A, B, C and D are points on the circle
Angle AOB = 140°
Angle OAC = 14°
Angle AOB = 2 × (Angle ACB) [angle subtended at the centre of the circle is twice the angle subtended at the circumference of the circle)
140° = 2 × (Angle ACB)
Angle ACB = (140°/2) = 70°
(Angle AOB) + (Angle OAB) + (Angle ABO) = 180° [sun of angles in a triangle is 180°]
But Angle OAB = Angle ABO = a [base angles of an iscosceles triangle are equal since OA and OB are both radii for the circle O)
(Angle AOB) + (Angle OAB) + (Angle ABO) = 180°
140° + a + a = 180°
2a = 40°
a = (40/2) = 20°
Angle OAB = Angle ABO = 20°
Angle CAB = (Angle OAC) + (Angle OAB) = 14° + 20° = 34°
Triangle ADC is an iscosceles triangle, hence,
Angle DAC = Angle ACD = x [base angles of an iscosceles triangle are equal]
But
(Angle DCB) + (Angle DAB) = 180° [Opposite angles of a cyclic quadilateral (a quadilateral inscribed in a circle) sum up to give 180°]
Angle DCB = (Angle DCA) + (Angle ACB) = (x + 70°)
Angle DAB = (Angle DAC) + (Angle CAB) = (x + 34°)
(Angle DCB) + (Angle DAB) = 180°
(x + 70°) + (x + 34°) = 180°
2x + 104° = 180°
2x = 180° - 104° = 76°
x = (76°/2) = 38°
Angle DAC = Angle ACD = 38°
Angle ACD = 38°
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