Questions: Solve the following problems. C) Given: triangle AKM inscribed in k(O, r=2), m angle A=33°, and O in AM Find: Perimeter of triangle AKM

Solve the following problems.
C) Given: triangle AKM inscribed in k(O, r=2), m angle A=33°, and O in AM
Find: Perimeter of triangle AKM
Transcript text: Solve the following problems. C) Given: $\triangle \mathrm{AKM}$ inscribed in $k(\mathrm{O}, r=2), \mathrm{m} \angle \mathrm{A}=33^{\circ}$, and $\mathrm{O} \in$ AM Find: Perimeter of $\triangle A K M$ Answer: $P_{A K M} \approx$ $\square$ units
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Solution

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Solution Steps

Step 1: Find the measure of arc KM

Since OE $\perp$ AM, AM is a chord and OE is a radius. The perpendicular from the center of a circle to a chord bisects the chord and its arc. Therefore, arc AE $\cong$ arc ME. Arc AM = 2 * $\angle$ A = 2 * 33° = 66°.

Step 2: Find the measure of arc AK

Since AAKM is inscribed in circle O, we know that the sum of the arcs intercepted by inscribed angles A and M is 180°. Arc KM + arc AK = 180°, so arc KM = $\angle$ 2A = 2 * 33° = 66°. Because arc KM = 66°, then arc AK = 180° - 66° = 114°.

Step 3: Find the length of chord AK

The length of a chord is given by the formula 2r * sin($\frac{c}{2}$) where $c$ is the central angle that subtends the chord. In this case, $c$ = 114° and $r$ = 2. Therefore AK = 2 * 2 * sin($\frac{114}{2}$) = 4 * sin(57°). AK $\approx$ 3.35

Step 4: Find AM

AM = 2 * AO * cos(A) AM = 2 * 2 * cos(33°) AM $\approx$ 3.35

Step 5: Find KM

Using the Law of Cosines where a, b, and c are sides of a triangle and angle C is opposite side c: $c^2 = a^2 + b^2 - 2ab$ cos(C) $KM^2 = AK^2 + AM^2 - 2(AK)(AM)$ cos(A) $KM^2 = 3.35^2 + 3.35^2 - 2(3.35)(3.35)$ cos(33°) $KM \approx$ 1.91

Final Answer: The perimeter of $\triangle$AKM is AK + KM + AM $\approx$ 3.35 + 3.35 + 1.91 $\approx$ 8.61 units.

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