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4 years ago

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Examine the diagram. It is not drawn to scale. A triangle has angles A, B, 80 degrees. The exterior angle to angle A is 150 degr

ees. Choose the statements that are true based on the diagram. ∠A is supplementary to the exterior angle. ∠A and the 80° angle are remote interior angles. 80° + m∠B + 150° = 360°. ∠B measures 70°. ∠A measures 40°.

2 answers:

Mnenie [13.5K]4 years ago

5 0

Answer:

A and D

Step-by-step explanation:

Just took it and got it right

german4 years ago

4 0

Answer:

∠A is supplementary to the exterior angle and ∠B measures 70°

Step-by-step explanation:I just answered this question and got it right.

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Answer:

Mother nature

Step-by-step explanation:

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3 years ago

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Pls help last one <br> Which one is it

Degger [83]

I think it’s SAS

SAS Means: If any two angles and the included side are the same in both triangles, then the triangles are congruent

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3 years ago

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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

strojnjashka [21]

Answer:

The missing side is $B = 6.0\ cm$

The missing angles are $\alpha = 56.2$ and $\theta = 93.8$

Step-by-step explanation:

Given

$A = 10\ cm$

$C = 12\ cm$

$\beta = 30$

The implication of this question is to solve for the missing side and the two missing angles

Represent

Angle A with $\alpha$

Angle B with $\beta$

Angle C with $\theta$

Calculating B

This will be calculated using cosine formula as thus;

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

Substitute values for A, C and $\beta$

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

$B^2 = 100 + 144 - 240 * 0.8660$

$B^2 = 100 + 144 - 207.8$

$B^2 = 36.2$

Take Square root of both sides

$B = \sqrt{36.2}$

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

Calculating $\alpha$

This will be calculated using cosine formula as thus;

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

Substitute values for A, B and C

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

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

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

Collect Like Terms

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

$-80 = -144Cos\alpha$

Divide both sides by -144

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

$0.5556 = Cos\alpha$

$\alpha = cos^{-1}(0.5556)$

$\alpha = 56.2$ <em>(Approximated)</em>

Calculating $\theta$

This will be calculated using cosine formula as thus;

$C^2 = B^2 + A^2 - 2BACos\theta$

Substitute values for A, B and C

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

$144 = 36 + 100 - 120Cos\theta$

Collect Like Terms

$144 - 36 - 100 = -120Cos\theta$

$8 = -120Cos\theta$

Divide both sides by -120

$\frac{8}{-120} = Cos\theta$

$-0.0667= Cos\theta$

$\theta = cos^{-1}(-0.0667)$

$\theta = 93.8$ <em>(Approximated)</em>

4 0

4 years ago

Include a step by step explanation pls ty

Eduardwww [97]

The division problem is 7 divided by 15. this can be written as 7/15. to do this, you first know that 15 doesn't go into 7, so you add a decimal point, and make the 7 70. 15 goes into 70 4 times. 15 times 4 is 60. 70-60 is 10. 15 doesn't go into 10, so you add another 0 and bring it down. 15 goes into 100 6 times. 15 times 6 is 90. 100-90 is 10. 15 doesn't go into 10, so you bring down another 0. 15 goes into 100 6 times. 15 times 6 is 90. 100-90 is 10. at this point, you can probably guess that the 6 will go on for a while. this means it is repeating. this means your answer is "a".

4 0

2 years ago

Type the correct answer in each box.

den301095 [7]

Answer:

m∠BPD: 120°

mBC + mAD =120°

Step-by-step explanation:

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3 years ago

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