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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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Use Lagrange multipliers to find the volume of the largest rectangular box in the first octant with three faces in the coordinat
tensa zangetsu [6.8K]

Answer:

The volume of the largest rectangular box (V) = 81/4

Step-by-step explanation:

<u>Step 1</u>:-

Given volume of the largest rectangular box in the first octant

V = l b h

let (x ,y, z) be the one vertex in the given plane

V = f(x, y, z) = x y z

given plane.   Ф (x, y, z) = x + 9y + 4z = 27 ........(1)

<u>Step( ii):</u>

By using  Lagrange multipliers

Suppose it is required to find the extreme for the function f(x, y, z) subject to the condition Ф (x, y, z) =0

Form Lagrange function F(x, y, z) = f(x, y, z) + λ  Ф (x, y, z) where λ is called the Lagrange multipliers

F(x, y, z) = x y z+ λ ( x + 9y + 4z - 27) ......(2)

Obtain the equations are δ F / δ x = 0

                                  δ f / δ x +λ  δ Ф  / δ x =0

                           ⇒ y z + λ ( 1) = 0

                          ⇒ y z = - λ ......(a)

Obtain the equations are δ F / δ y = 0

                            δ f / δ x +λ  δ Ф  / δ x =0

                        ⇒ x z + λ ( 9) = 0

                        ⇒ \frac{xz}{9} = - λ .......(b)

Obtain the equations are δ F / δ z = 0

                                  δ f / δ x +λ  δ Ф  / δ z =0

                                   x y + λ ( 4) = 0

                                    ⇒ \frac{xy}{4} = - λ .......(c)

<u>Step (iii):-</u>

Equating (a) and (b) equations

we get         y z = \frac{xz}{9}

cancel 'z' value we get  x = 9y  .......(d)

Equating (b) and (c) equations

we get     \frac{xy}{4}  = \frac{xz}{9}

cancel 'x' value on both sides , we get

              4z =9y  ......(e)

substitute (d) (e) values in equation (1)  we get

x + 9y + 4z -27 = 0

9y + 9y + 9y -27 =0

27y -27 =0

<u> y = 1</u>

substitute y =1 in x = 9y

<u>x = 9</u>

substitute y =1 in 4z =9y

<u>z = 9 /4</u>

therefore the dimensions are x =9 , y=1 and z = 9 /4

<u>Conclusion</u>:-

The largest volume of the rectangular box V = x y z

substitute x =9 , y=1 and z = 9 /4

V = 9(1)(9/4) =81/4

The largest volume of the rectangular box (V) = 81/4

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substitute x =9 , y=1 and z = 9 /4

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