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3 years ago

A regular polygon has a measure of (8x+6) and a measure of (4x+12) find the measure of each angle please show work

Mathematics

1 answer:

stepladder [879]3 years ago

6 0

Answer:

The measure of each angle is 18°

Step-by-step explanation:

we know that

A regular polygon is a polygon whose all sides are equal and also all angles are equal.

In this problem

$(8x+6)=(4x+12)$

Solve for x

$(8x-4x)=(12-6)$

$4x=6$

$x=1.5$

Substitute the value of x

$(8x+6)=8(1.5)+6=18\°$

therefore

The measure of each angle is 18°

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3 years ago

Tan9 - tan27 - tan63 + tan 81 = ?

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3 years ago

2 of a dozen eggs are cracked. What percent of the carton is NOT cracked using the percent proportion

levacccp [35]

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6 0

3 years ago

The measure of an angle is 2.8.what is the measure of its complementary angle

german

Answer:

87.2º

Step-by-step explanation:

Complimentary angles total 90º

c + 2.8 = 90

c = 90 - 2.8

c = 87.2º

3 0

3 years ago

Read 2 more answers

Please prove this........

Crazy boy [7]

Answer: see proof below

Step-by-step explanation:

Given: A + B + C = π → C = π - (A + B)

→ sin C = sin(π - (A + B)) cos C = sin(π - (A + B))

→ sin C = sin (A + B) cos C = - cos(A + B)

Use the following Sum to Product Identity:

sin A + sin B = 2 cos[(A + B)/2] · sin [(A - B)/2]

cos A + cos B = 2 cos[(A + B)/2] · cos [(A - B)/2]

Use the following Double Angle Identity:

sin 2A = 2 sin A · cos A

Proof LHS → RHS

LHS: (sin 2A + sin 2B) + sin 2C

$\text{Sum to Product:}\qquad 2\sin\bigg(\dfrac{2A+2B}{2}\bigg)\cdot \cos \bigg(\dfrac{2A - 2B}{2}\bigg)-\sin 2C$

$\text{Double Angle:}\qquad 2\sin\bigg(\dfrac{2A+2B}{2}\bigg)\cdot \cos \bigg(\dfrac{2A - 2B}{2}\bigg)-2\sin C\cdot \cos C$

$\text{Simplify:}\qquad \qquad 2\sin (A + B)\cdot \cos (A - B)-2\sin C\cdot \cos C$

$\text{Given:}\qquad \qquad \quad 2\sin C\cdot \cos (A - B)+2\sin C\cdot \cos (A+B)$

$\text{Factor:}\qquad \qquad \qquad 2\sin C\cdot [\cos (A-B)+\cos (A+B)]$

$\text{Sum to Product:}\qquad 2\sin C\cdot 2\cos A\cdot \cos B$

$\text{Simplify:}\qquad \qquad 4\cos A\cdot \cos B \cdot \sin C$

LHS = RHS: 4 cos A · cos B · sin C = 4 cos A · cos B · sin C $\checkmark$

7 0

3 years ago