Question

I forgot how to do this, can someone please help

Answer 1

Solution given:

<R=<J=49°[ inscribed angle standing on a same Aaka equal in a circle]

<K=90°[inscribed is triangle on a semicircle is 90°]

now in triangle JKL

<J+<K+<L=180°[ sum of interior angles of a triangle is 180°]

49°+90°+<L=180°

<L=180°-49°-90°=41°

:.arcJK=41°

is your answer.

Answer 2

Question:

Find the measure of the arc or angle indicated.

Solution:

In ∆LRK & ∆JKL,

∠KRL= ∠LJK = 49°

• ( inscribed angle standing on a same Aaka equal in a circle)

Now , In ∆JKL,

∠JKL = 90°

• (angle suspended to the center of the circle =90°)

Now,

• construction: A line segment OK joining the points O(center)& K

Since, OK&OJ are the radii of the same circle,

Hence, OK = OJ

.°. ∠JKO = ∠OJK = 49°

Now,

In ∆ JOK,

∠JKO +∠OJK+ ∠JOK = 180°

• Angle sum property of a∆

∠JOK = 180°- (∠JKO +∠OJK)

∠JOK = 180°-(49°+49°)

∠JOK = 180°- 98°

∠JOK = 82°

Hence, 82° is your answer

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Hope it helps

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