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"angle CBD" should be "angle BCD"
Problem. Two circles of the same radius intersect at the angle 2d/3. Express in degrees the smaller of the arcs contained between the intersection points.
Can I get some clarification please? There was a remark appended to the problem which clarified the problem a bit, but I'm still not completely certain about what is required to be found, nor what is given.
Thanks
Here is an equivalent question: Two circles of the same radius intersect at 60 degree angle(s). In what proportion do they divide each other?
[In fact there is some ambiguity in this problem, for - which of the angles between two tangents is 60 degrees?]
258. Compute the degree measure of an inscribed angle intercepting an arc congruent to 1/12th part of the circle.
An arc congruent to 1/12th part of a circle has 30 circular degrees, so an inscribed angle intercepting the arc would have half as many angular degrees, so 15 angular degrees.
259. A disk is partitioned into two disk segments by a chord dividing the circle in the proportion 5:7. Compute the angles enclosed by these segments.
Since the circle is divided in the proportion, one arc will contain 5/12th of the circle and the other, 7/12th, so the disk segments contain 150 and 210 circular degrees.
260. Two chords intersect at an angle 36°15'30". Express in degrees, minutes, and seconds the two arcs intercepted by this angle and the angle vertical to it, if one of these arcs measures 2/3 of the other.
Then, using 126(1), 36°15'30" is the semisum of the two arcs, so 72°31' is the sum of the two arcs. Let x be one of the measures of an arc and (2/3)x is the other, then x=72°31'(3/5)=43°30'36", and the other arc has measure (2/3)43°30'36"=29°0'24".
261. The angle between two tangents drawn from the same point to a circle is 25°15'. Compute the arcs contained between the tangency points.
Let's extend the tangents and the perpendicular radii to intersect. Then, we get four right triangles where the given angle is an acute angle, ie the given angle is the vertical angle between the extended perpendicular radii (not the angle we want). Since the radii intercept at the center, the measure of the arc corresponds to the angular degrees in the supplement to the given angle, ie 154°45'.
262. Compute the angle formed by a tangent and a chord, if the chord divides the circle in the proportion 3:7.
Then, the chord divides the arcs with measures of (3/10)360°=108° and 252°. Using §125, the angle formed by the chord and tangent will measure half the intercepted arc, so 54° and 126°.
263. Two circles of the same radius intersect at the angle 2d/3. Express in degrees the smaller of the arcs contained between the intersection points.
Since the circles have the same radius, the common chord and line of centers are axes of symmetry that are perpendicular to each other. Thus, the quadrilateral formed by extending the tangents at the intersection points must be a rhombus and thus, a parallelogram, so the tangents are pairwise parallel.
Let x be the angle between radii going to one intersection point. For the tangents of one circle, extend them to be secants of the other circle, ABE and CBD, where E and C are the intersection points. Then, applying §125, the arc ACE and CED both have measures of 2x, where ACE=AC+CE and CED=CE+ED, so ACE+CED=4x=2CE+AC+ED. Since we have parallel tangents, ∠DBE and ∠ABC are both congruent to x, so applying §126(1) to either, the sum of the arcs ED and AC have a measure of 2x, ie AC+ED=2x, thus 2CE=2x, so CE=x. Now, x is either 2d/3 or 4d/3, depending on which angle between the tangent lines the given angle is referring to.
264. A tangent is drawn through one endpoint of a diameter and a secant through the other, so that they make the angle 20°30'. Compute the smaller of the arcs contained between the tangent and the secant.
We can't use §126(2) because one of the sides of the angle is a tangent and doesn't intersect the circle, but we can expand the theorem using basically the same proof. Let AB be the tangent and BDC be the secant where B is in the exterior of the disk. Then, again, ∠ADC is an exterior angle to △ABD, so ∠ABC+∠DAB=∠ADC so ∠ABC=∠ADC-∠DAB, where ∠ADC is an inscribed angle intercepting the diameter and ∠DAB is an angle formed by a chord and a tangent. Then, using §124(2) and §125, ∠ADC contains 90° and ∠DAB measures a half of the arc, AD which is what we want to compute in this exercise. We know ∠ABC=20°30', thus ∠DAB=69°30' so the arc AD contains 139°.
Alternatively, notice the tangent, secant, and diameter form a right triangle, where the arc we want is intercepted by the complement to the given angle, which again, results in 2(90°-20°30')=139°.
265. The feet of the perpendiculars dropped from a given point A to lines passing through another given point B.
The geometric locus is the circle with diameter AB centered between A and B, ie imagine Figure 138, the perpendicular and the corresponding line basically create a right inscribed angle.
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