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2 years ago

5

I need help plz thx

1 answer:

slamgirl [31]2 years ago

5 0

Answer:

Perimeter = 36 units

Step-by-step explanation:

The figure given has been labelled with letters for easy reference.

Recall: when two tangents segments meet at a point outside the circle, the two segments that are tangent to the circle are congruent to each.

Perimeter = a + b + c + d + e + f

e = 4 (given)

e = f = 4 (tangents drawn from an external point)

b = 10 - f

Substitute

b = 10 - 4

b = 6

b = a = 6 (tangents drawn from an external point)

c = 8 (given)

c = d = 8 (tangents drawn from an external point)

Perimeter = 6 + 6 + 8 + 8 + 4 + 4

Perimeter = 36 units

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Solve for x (please help i dont know how to this one ;-;)

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Use this app is the best app ever

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2 years ago

Solve the oblique triangle where side a has length 10 cm, side c has length 12 cm, and angle beta has measure thirty degrees. Ro

Ugo [173]

Answer:

$Side\ B = 6.0$

$\alpha = 56.3$

$\theta = 93.7$

Step-by-step explanation:

Given

Let the three sides be represented with A, B, C

Let the angles be represented with $\alpha, \beta, \theta$

[See Attachment for Triangle]

$A = 10cm$

$C = 12cm$

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What the question is to calculate the third length (Side B) and the other 2 angles ( $\alpha\ and\ \theta$ )

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When two angles of a triangle are known, the third side is calculated as thus;

$B^2 = A^2 + C^2 - 2ABCos\beta$

Substitute: $A = 10$ , $C =12$ ; $\beta = 30$

$B^2 = 10^2 + 12^2 - 2 * 10 * 12 *Cos30$

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$B^2 = 100 + 144 - 207.846096907$

$B^2 = 36.153903093$

Take Square root of both sides

$\sqrt{B^2} = \sqrt{36.153903093}$

$B = \sqrt{36.153903093}$

$B = 6.0128115797$

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Calculating Angle $\alpha$

$A^2 = B^2 + C^2 - 2BCCos\alpha$

Substitute: $A = 10$ , $C =12$ ; $B = 6$

$10^2 = 6^2 + 12^2 - 2 * 6 * 12 *Cos\alpha$

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$100 = 36 + 144 - 144 *Cos\alpha$

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Subtract 180 from both sides

$100 - 180 = 180 - 180 - 144 *Cos\alpha$

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Divide both sides by -144

$\frac{-80}{-144} = \frac{- 144 *Cos\alpha}{-144}$

$\frac{-80}{-144} = Cos\alpha$

$0.5555556 = Cos\alpha$

Take arccos of both sides

$Cos^{-1}(0.5555556) = Cos^{-1}(Cos\alpha)$

$Cos^{-1}(0.5555556) = \alpha$

$56.25098078 = \alpha$

$\alpha = 56.3$ <em>(Approximated)</em>

Calculating $\theta$

Sum of angles in a triangle = 180

Hence;

$\alpha + \beta + \theta = 180$

$30 + 56.3 + \theta = 180$

$86.3 + \theta = 180$

Make $\theta$ the subject of formula

$\theta = 180 - 86.3$

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The center of a circle is at the origin on a coordinate grid. A line with a positive slope intersects the circle at (0, 7). Whic

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The true statement is the circle has the radius equal to 7.

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As, standard form of equation of circle is

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Answer:

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