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Ivahew [28]
3 years ago
5

SOMEONE PLEASE HELP ME! I have a few of the explanations but I'm not sure if I need more please suggest some thank you

Mathematics
1 answer:
lesantik [10]3 years ago
5 0

Answer:

For tingle #1

We can find angle C using the triangle sum theorem: the three interior angles of any triangle add up to 180 degrees. Since we know the measures of angles A and B, we can find C.

C=180-(A+B)

C=180-(21.24+27.14)

C=131.62

We cannot find any of the sides. Since there is noting to show us size, there is simply just not enough information; we need at least one side to use the rule of sines and find the other ones. Also, since there is nothing showing us size, each side can have more than one value.  

For triangle #2

In this one, we can find everything and there is one one value for each.

- We can find side c

Since we have a right triangle, we can find side c using the Pythagorean theorem

b^2=a^2+c^2

4^2=2^2+c^2

16=4+c^2

12=c^2

c=\sqrt{12}

c=2\sqrt{3}

- We can find angle C using the cosine trig identity

cos(C)=\frac{adjacent}{hypotenuse}

cos(C)=\frac{2}{4}

C=arccos(\frac{2}{4} )

C=60

- Now we can find angle A using the triangle sum theorem

A=180-(B+C)

A=180-(90+60)

A=30

For triangle #3

Again, we can find everything and there is one one value for each.

- We can find angle A using the triangle sum theorem

A=180-(B+C)

A=180-(90+34.88)

A=55.12

- We can find side a using the tangent trig identity

tan(C)=\frac{opposite-side}{adjacent-side}

tan(34.88)=\frac{7}{a}

a=\frac{7}{tan(34.88)}

a=10.04

- Now we can find side b using the Pythagorean theorem

b^2=a^2+c^2

b^2=10.04^2+7^2

b^2=149.8

b=\sqrt{149.8}

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&A=2\left[2 h^{3 / 2} \sqrt{p}-\frac{8(\sqrt{p})^{3}(\sqrt{h})^{3}}{12 p}-0\right] \\\\&A=2\left[2 h^{k / 2} \sqrt{p}-\frac{2}{3} \sqrt{p} \cdot h^{3 / 2}\right] \\\\&A=2 \times \frac{4 \sqrt{p} \cdot h^{3 / 2}}{3} \\\\&A=\frac{8}{3} \cdot \sqrt{p} \cdot \sqrt{h} \cdot h \\\\&A=\frac{8}{3} \sqrt{p h} \cdot h

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A=\frac{8}{3} \sqrt{p h} \cdot h

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