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zvonat [6]
2 years ago
13

Find the measure of ∠EGC. Circle A with chords EF and CD that intersect at point G, the measure of arc EC is 50 degrees, the mea

sure of angle EGC is 7x plus 7 degrees, and the measure of arc DF is 10x degrees. 140° 90° 70° 50°
Mathematics
2 answers:
miv72 [106K]2 years ago
8 0

Answer:

m∠EGC=70°

Step-by-step explanation:

we know that

The measure of the inner angle is the semi-sum of the arcs comprising it and its opposite

so

m∠EGC=(1/2)[arc EC+arc DF]

<u><em>Find the value of x</em></u>

we have

m∠EGC=(7x+7)°

arc EC=50°

arc DF=10x°

substitute and solve for x

(7x+7)°=(1/2)[50°+10x°]

14x+14=50+10x

14x-10x=50-14

4x=36

x=9

<u><em>Find the measure of angle EGC</em></u>

m∠EGC=(7x+7)°

substitute the value of x

m∠EGC=(7(9)+7)°=70°

loris [4]2 years ago
7 0

Answer:

C. 70

Step-by-step explanation:

Got it on my test.

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Lucia knows the fourth term in a sequence is 55 and the ninth term in the same sequence is 90. Explain how she can find the comm
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2 years ago
How many eggs are needed to make 1 dozen waffles, assuming you have enough of all other ingredients? given: 2 cups flour + 3 egg
malfutka [58]

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D is the correct option.

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2 years ago
Solve: 5x(8-4) divide 4-2=
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5x(4)/4 - 2

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8 0
2 years ago
Given: KL ║ NM , LM = 45, m∠M = 50° KN ⊥ NM , NL ⊥ LM Find: KN and KL
Mice21 [21]

Answer:

KL=45\tan 50^{\circ}\sin 50^{\circ}\approx 41.08\\ \\KN=45\sin 50^{\circ}\approx 34.47

Step-by-step explanation:

Given:

KL ║ NM ,

LM = 45

m∠M = 50°

KN ⊥ NM  

NL ⊥ LM

Find: KN and KL

1. Consider triangle NLM. This is a right triangle, because NL ⊥ LM. In this triangle,

LM = 45

m∠M = 50°

So,

\tan \angle M=\dfrac{\text{opposite leg}}{\text{adjacent leg}}=\dfrac{NL}{LM}=\dfrac{NL}{45}\\ \\NL=45\tan 50^{\circ}

Also

m\angle LNM=90^{\circ}-50^{\circ}=40^{\circ} (angles LNM and M are complementary).

2. Consider triangle NKL. This is a right triangle, because KN ⊥ NM . In this triangle,

NL=45\tan 50^{\circ}

m\angle KLN=m\angle LNM=40^{\circ} (alternate interior angles)

m\angle KNL=90^{\circ}-40^{\circ}=50^{\circ} (angles KNL and KLN are complementary).

So,

\sin \angle KNL=\dfrac{\text{opposite leg}}{\text{hypotenuse}}=\dfrac{KL}{LN}=\dfrac{KL}{45\tan 50^{\circ}}\\ \\KL=45\tan 50^{\circ}\sin 50^{\circ}\approx 41.08

and

\cos \angle KNL=\dfrac{\text{adjacent leg}}{\text{hypotenuse}}=\dfrac{KN}{LN}=\dfrac{KN}{45\tan 50^{\circ}}\\ \\KN=45\tan 50^{\circ}\cos 50^{\circ}=45\sin 50^{\circ}\approx 34.47

3 0
2 years ago
Read 2 more answers
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