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Naddik [55]
2 years ago
13

Given the angles inscribed in the circle, find m ∠ DBC.

Mathematics
2 answers:
Cloud [144]2 years ago
6 0

Answer:

m∠DBC = 51°

Step-by-step explanation:

<u>Circle Theorem vocabulary</u>

  • Chord: A straight line joining two points on the circle.
  • Inscribed angle: The angle formed when two chords meet at one point on a circle.
  • Intercepted arc: The arc that is between the endpoints of the chords that form the inscribed angle.

<u>Inscribed Angle Theorem</u>

The measure of an <u>inscribed angle</u> is half the measure of the <u>intercepted arc</u>.

Therefore:

\sf \implies \angle DBC=\dfrac{1}{2}\:\overset{\frown}{BC}

\sf \implies \angle DBC=\dfrac{1}{2}\:(102^{\circ})

\sf \implies \angle DBC=51^{\circ}

Learn more about the Inscribed Angle Theorem here:

brainly.com/question/27934538

brainly.com/question/27943630

ASHA 777 [7]2 years ago
5 0

\huge\underline{\underline{\boxed{\mathbb {SOLUTION:}}}}

To determine ∠DBC, we will apply the theorem relating the inscribed angle to the intercepted arc:

\longrightarrow \sf{Intercepted \: arc= (inscribe \: angle)}

\longrightarrow \sf{Intercept \: arc= 102^\circ}

\longrightarrow \sf{Inscribed \: angle= \angle DBC }

\longrightarrow \sf{102=2( \angle DBC)}

\longrightarrow \sf{\angle DBC= \dfrac{102}{2} }

▪ \longrightarrow \sf{\angle DBC = 51^\circ}

\huge\underline{\underline{\boxed{\mathbb {ANSWER:}}}}

\boxed{\bf{\large m \: \angle DBC= 51^\circ}}

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If a = 7 and b = 11, what is the measure of ∠B? (round to the nearest tenth of a degree) A) 32.5° B) 39.2° C) 50.5° D) 57.5°
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D) 57.5°

Step-by-step explanation:

As the question is not complete. So, let's suppose it is a right angle triangle then, we can apply Pythagoras theorem to calculate the hypotenuse or the third side.

Pythagoras Theorem = c^{2} = a^{2} + b^{2}

a = 7 and b = 11

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Plugging in the values, we will get:

c^{2} = 49 + 121

c^{2} = 170

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Law of sine = \frac{a}{sinA} = \frac{b}{sinB} = \frac{c}{sinC}

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\frac{c}{sinC} = \frac{b}{sinB}

\frac{\sqrt{170} }{sin90} =  \frac{11}{sinB}

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sinB = \frac{11}{\sqrt{170} }

B = sin^{-1} (\frac{11}{\sqrt{170} })

B = 57.5°

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