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3 years ago

Solve for the value of x

Mathematics

2 answers:

aev [14]3 years ago

7 0

Answer:

x = 2

Step-by-step explanation:

Given 2 intersecting chords in a circle, then

The product of the parts of one chord is equal to the product of the parts of the other chord , that is

12x = 6(x + 2) = 6x + 12 ( subtract 6x from both sides )

6x = 12 ( divide both sides by 6 )

x = 2

Mariulka [41]3 years ago

3 0

Answer:

$x=2$

Step-by-step explanation:

The term 'secant' refers to a line segment that intersects a circle in two places. When two secants intersect inside a circle, one can use the product of lengths theory to form ratios between the parts of intersection between the secants. This ratio can be described as the following, let ( $secant_A$ ) and ( $secant_B$ ) represent the two secants in the circle. Part (1) and (2) will refer to the two parts formed after the intersection of the secants.

$(secant_A_1)(secant_A_2)=(secant_B_1)(secant_B_2)$

Use this formula in the given situation, substitute the given values in and solve for the unknown,

$(secant_A_1)(secant_A_2)=(secant_B_1)(secant_B_2)$

$(6)(x+2)=(12)(x)$

Simplify,

$(6)(x+2)=(12)(x)$

$6x+12=12x$

$12=6x$

$x=2$

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3 years ago

Cosθ=−2√3 , where π≤θ≤3π2 .

Alex787 [66]

Answer:

$sin(\theta + \beta) = -\frac{\sqrt{7}}{5}-4\frac{\sqrt{2}}{15}$

Step-by-step explanation:

step 1

Find the $sin(\theta)$

we know that

Applying the trigonometric identity

$sin^2(\theta)+ cos^2(\theta)=1$

we have

$cos(\theta)=-\frac{\sqrt{2}}{3}$

substitute

$sin^2(\theta)+ (-\frac{\sqrt{2}}{3})^2=1$

$sin^2(\theta)+ \frac{2}{9}=1$

$sin^2(\theta)=1- \frac{2}{9}$

$sin^2(\theta)= \frac{7}{9}$

$sin(\theta)=\pm\frac{\sqrt{7}}{3}$

Remember that

π≤θ≤3π/2

Angle θ belong to the III Quadrant

That means ----> The sin(θ) is negative

$sin(\theta)=-\frac{\sqrt{7}}{3}$

step 2

Find the sec(β)

Applying the trigonometric identity

$tan^2(\beta)+1= sec^2(\beta)$

we have

$tan(\beta)=\frac{4}{3}$

substitute

$(\frac{4}{3})^2+1= sec^2(\beta)$

$\frac{16}{9}+1= sec^2(\beta)$

$sec^2(\beta)=\frac{25}{9}$

$sec(\beta)=\pm\frac{5}{3}$

we know

0≤β≤π/2 ----> II Quadrant

sec(β), sin(β) and cos(β) are positive

$sec(\beta)=\frac{5}{3}$

Remember that

$sec(\beta)=\frac{1}{cos(\beta)}$

therefore

$cos(\beta)=\frac{3}{5}$

step 3

Find the sin(β)

we know that

$tan(\beta)=\frac{sin(\beta)}{cos(\beta)}$

we have

$tan(\beta)=\frac{4}{3}$

$cos(\beta)=\frac{3}{5}$

substitute

$(4/3)=\frac{sin(\beta)}{(3/5)}$

therefore

$sin(\beta)=\frac{4}{5}$

step 4

Find sin(θ+β)

we know that

$sin(A + B) = sin A cos B + cos A sin B$

In this problem

$sin(\theta + \beta) = sin(\theta)cos(\beta)+ cos(\theta)sin (\beta)$

we have

$sin(\theta)=-\frac{\sqrt{7}}{3}$

$cos(\theta)=-\frac{\sqrt{2}}{3}$

$sin(\beta)=\frac{4}{5}$

$cos(\beta)=\frac{3}{5}$

substitute the given values in the formula

$sin(\theta + \beta) = (-\frac{\sqrt{7}}{3})(\frac{3}{5})+ (-\frac{\sqrt{2}}{3})(\frac{4}{5})$

$sin(\theta + \beta) = (-3\frac{\sqrt{7}}{15})+ (-4\frac{\sqrt{2}}{15})$

$sin(\theta + \beta) = -\frac{\sqrt{7}}{5}-4\frac{\sqrt{2}}{15}$

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4 years ago