Answer:
3. d = 90°, e = 5, f = 7
4. x = 50°, y = 90°; x = 22°, y = 136°; x = y = 35°
5. a = b = √91; a = √33, b = 2√33
6. b = 2√14
Explanation:
You want the solutions to multiple problems involving chords and related triangles.
Relations
What you need to know here is that the line from the center of a chord through the center of the circle is the perpendicular bisector of the chord.
In 3a, this means angle d° is 90°. In 3b, this means e = 5, since the two halves of the chord have equal measures.
The triangle formed with a chord and radii, as in 4b and 4c is an isosceles triangle. The base angles are equal, and the sum of angles is 180°, as for any triangle.
In problem 6, PQ passes through the center of the circle, so is a diameter, PO is a radius, so is half the total length. Side PS in that triangle will have the same length as SR (also called b).
The Pythagorean theorem is used for problems 5 and 6.
3. Angle and halves
As discussed, the angle at the chord is a right angle.
d = 90°
The two halves of the chord are equal.
e = 5
Each half of the chord is half the whole length.
f = 14/2 = 7
4. Triangles
(a) ∆ACO is a right triangle, so y = 90°, and x is the complement of 40°.
x = 90° -40° = 50°
y = 90°
(b) Base angles are equal, and the central angle is the supplement of their sum.
x = 22°
y = 180° -2(22°) = 136°
(c) Base angles are equal and each is half the supplement of the central angle.
x = y = (180° -110°)/2 = 35°
5. Right triangles
(a) The Pythagorean relation applies.
a = b = √(10² -3²) = √91
(b) Same here.
a = √(7² -4²) = √33
b = 2a = 2√33
6. Another right triangle
The radius is half the diameter, so the hypotenuse (PO) of the triangle is 18/2 = 9. The unknown chord half (PS) is equal in length to the half marked b, so ...
b = √(9² -5²) = √56
b = 2√14
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Additional comment
We assume you're familiar with the Pythagorean relation. For right triangle legs a, b, and hypotenuse c, you have ...
a² +b² = c²
Solving this for an unknown leg, you find ...
b = √(c² -a²)
This is the form we used in 5b and 6.
The hash marks on the chord in 3a, 3c and 4a identify congruent segments. They are telling you the line to the circle center is dividing the chord in half, hence is perpendicular to the chord. (Elsewhere, the right angles are all marked.)
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