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3 years ago

A triangle has sides of length 2, 2, and 3. Which of the following is the measure of the obtuse angle of the triangle?

Mathematics

1 answer:

kramer3 years ago

7 0

Answer:

$97.18^{\circ}$

Step-by-step explanation:

An obtuse angle in the triangle is opposite to the greatest side. The greatest side in the triangle with sides lengths 2, 2 and 3 is side with length 3. By the cosine rule,

$3^2=2^2+2^2-2\cdot 2\cdot 2\cdot \cos \alpha,\\ \\9=4+4-8\cos \alpha,\\ \\-8\cos \alpha=9-4-4=1,\\ \\\cos \alpha=-\dfrac{1}{8},\\ \\\alpha\approx 97.18^{\circ}.$

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Answer: i’m not sure if it’s just simplifying or factoring. but for simplifying it equals. 1/8 or .125

Explanation: calculator

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2 years ago

Read 2 more answers

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]

$A = 10cm$

$C = 12cm$

$\beta = 30$

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

Solving for Side B;

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

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

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

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

$B^2 = 100 + 144 - 240*0.86602540378$

$B^2 = 100 + 144 - 207.846096907$

$B^2 = 36.153903093$

Take Square root of both sides

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

$B = \sqrt{36.153903093}$

$B = 6.0128115797$

$B = 6.0$ <em>(Approximated)</em>

Calculating Angle $\alpha$

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

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

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

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

$100 = 180 - 144 *Cos\alpha$

Subtract 180 from both sides

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

$-80 = - 144 *Cos\alpha$

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

$\theta = 180 - 86.3$

$\theta = 93.7$

5 0

2 years ago

Find the area of the trapezoid. points are (-5,-3)(4,-3)(6,-7)(-7,-7)

Olegator [25]

Answer:

44 square units

Step-by-step explanation:

The area of a trapezoid with bases b₁ and b₂ and height h is given by the formula

$A=\left(\dfrac{b_1+b_2}{2}\right)h$

If you're wondering how we get this formula, check the attached illustration (remember the area of a parallelogram is its base multiplied by its height)! Moving on to our trapezoid, the pairs of points (-5,-3)(4,-3) and (6,-7)(-7,-7) form two horizontal segments, which form b₁ and b₂, and our height is the distance between the y-coordinates -3 and -7, which is 4. We can find b₁ and b₂ by finding the distance between the x coordinates in their pairs of points:

$b_1=|-5-4|=|-9|=9\\b_2=|6-(-7)|=|6+7|=13$

Putting it altogether:

$A=\left(\dfrac{9+13}{2}\right)(4)=\left(\dfrac{22}{2}\right)(4)=(11)(4)=44$

So the area of our trapezoid is 44.

4 0

2 years ago

A drawing has a scale of 3 in. : 2 ft. What is the scale factor of the drawing?

icang [17]

The scale factor of the drawing as described is; 1.5in/ft

<h3>Scale factor of drawings</h3>

According to the question;

The given scale factor of the drawing is; 3 in: 2 ft.

In essence, 3 inches on the drawing board represents 2 ft of the object.

Hence, by finding the quotient of the units, we have;

3in/2ft = 1.5 in/ft