Given: ∆ABC, m∠A = 35°, Circle k(O, r=3), O∈ AB, AB is a diameter in the circle passing through point O. AC and CB are chords which intersect. Not stated that <ACB is 90 degrees.
Find: Perimeter of ∆ABC
Thank you
angle ACB is in fact 90°, since AB is a diameter.
So the perimeter is
p = AB + CB + CA
= 6 + 6sin35° + 6sin55°
To find the perimeter of triangle ABC, we first need to find the lengths of the sides.
Since AB is a diameter of circle k passing through O, it is the longest chord and splits the circle into two equal halves. Therefore, the length of AB is equal to twice the radius of the circle.
Given r = 3, the length of AB = 2 * 3 = 6 units.
Now, let's find the lengths of AC and CB.
Since AC and CB are chords of circle k, and they intersect, the point of intersection must be on the circle. Let's call this point P.
Since O is on AB and AB is a diameter, we can conclude that ∆BAO is a right triangle with a right angle at O.
Therefore, ∠AOB = 90°
Since ∠A = 35°, we can find ∠B:
∠A + ∠B + ∠C = 180° (Sum of angles in a triangle)
35° + ∠B + ∠C = 180°
∠B + ∠C = 180° - 35°
∠B + ∠C = 145°
Since ∠AOB = 90°, ∠COP = 180° - ∠AOB = 180° - 90° = 90°. Therefore, ∠COP is a right angle.
Since the opposite angles in a cyclic quadrilateral are supplementary, we have:
∠COP + ∠B = 180°
Substituting ∠COP = 90°:
90° + ∠B = 180°
∠B = 180° - 90°
∠B = 90°
So, ∠A = 35° and ∠B = 90°.
Since ∠B is a right angle, we can conclude that ∠ACB is also a right angle.
Therefore, triangle ABC is a right triangle with sides AB = 6, AC, and BC.
Using the Pythagorean theorem:
AC² + BC² = AB²
AC² + BC² = 6²
AC² + BC² = 36
Now, let's solve for AC and BC.
We can use the fact that the product of the segments of intersecting chords is equal.
So, AC * CB = PC * CP
Using the lengths PC = CP = r = 3:
AC * CB = 3 * 3
AC * CB = 9
Now, we have two equations:
AC * CB = 9
AC² + BC² = 36
Solving these equations simultaneously will give us the lengths of AC and BC.
Let's rearrange the first equation to solve for AC:
AC = 9 / CB
Substituting this into the second equation:
(9 / CB)² + CB² = 36
81 / CB² + CB² = 36
81 + CB⁴ = 36 * CB²
CB⁴ - 36 * CB² + 81 = 0
This is a quadratic equation in terms of CB². We can solve this equation using the quadratic formula:
CB² = (-b ± √(b² - 4ac)) / 2a
In this case, a = 1, b = -36, and c = 81. Substituting these values into the quadratic formula:
CB² = (-(-36) ± √((-36)² - 4 * 1 * 81)) / 2 * 1
CB² = (36 ± √(1296 - 324)) / 2
CB² = (36 ± √972) / 2
CB² = (36 ± √(324 * 3)) / 2
CB² = (36 ± 18√3) / 2
CB² = 18 ± 9√3
To find the values of AC and BC, we need to take the square root of CB²:
CB = √(18 ± 9√3)
Since AC = 9 / CB:
AC = 9 / √(18 ± 9√3)
Therefore, the perimeter of triangle ABC is:
Perimeter = AB + AC + BC
Perimeter = 6 + (9 / √(18 ± 9√3)) + √(18 ± 9√3)
To find the perimeter of ∆ABC, we need to determine the lengths of its three sides.
We know that AB is a diameter of the circle, passing through point O (the center of the circle). This means that AB is twice the radius of the circle, which is 3 units, so AB = 2 * 3 = 6 units.
Next, we need to find the lengths of AC and CB. Since AC and CB are chords that intersect, we can draw a line segment from point O to the point of intersection, creating two right triangles: ∆AOC and ∆COB.
Since AB is a diameter, ∠AOB is a right angle. Since O∈ AB and AC and CB are chords that intersect, OC is a height of ∆COB and OA is a height of ∆AOC. Therefore, ∠AOC and ∠BOC are right angles as well.
Since ∆AOC and ∆BOC are right triangles and the hypotenuse of both triangles is the same (AB), they are congruent.
So, we can determine the lengths of AC and CB by using the Pythagorean theorem in either triangle. Let's use ∆AOC:
In ∆AOC, we have ∠AOC = 90° and m∠A = 35°. Therefore, m∠OAC = 90° - 35° = 55°.
Now, using trigonometry, we can determine the length of AC:
AC = AO * tan(m∠OAC)
We know that AO is the radius of the circle, which is 3 units. So:
AC = 3 * tan(55°)
Now, use a calculator to find the value of tan(55°) and multiply it by 3 to get the length of AC.
Similarly, we can find the length of CB using ∆BOC.
Once we have the lengths of AB, AC, and CB, we can calculate the perimeter of ∆ABC by adding the lengths of its three sides:
Perimeter = AB + AC + CB