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4 years ago

Lines t and s (picture attached)

Mathematics

1 answer:

KatRina [158]4 years ago

8 0

1 and 3 or 2 and 4. The correct answer is B.

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A surveyor is measuring the distance across a small lake. He has set up his transit on one side of the lake 130 feet from a pili

Dimas [21]

Answer:

x = 225.16 feet

Step-by-step explanation:

Given that,

The angle between the piling and the pier is 60°.

A surveyor has set up his transit on one side of the lake 130 feet from a piling that is directly across from a pier on the other side of the lake.

We need to find the distance between the piling and the pier. Let the distance be d. Using the trigonometry to find it as follows :

$\tan(60)=\dfrac{x}{130}\\\\x=\tan(60)\times 130\\\\x=225.16\ feet$

So, the required distance is 225.16 feet.

4 0

3 years ago

Can someone please help me out

ira [324]

$BD \: is \: the \: bisector \: of \: \widehat{ABC} \Rightarrow \widehat{ABD}=\widehat{CBD} \\\Rightarrow \widehat{ABC} = \widehat{ABD} + \widehat{CBD} = 2\widehat{ABD} = 2 \times 39 = 78° \\ \Rightarrow D. \: 78°$

4 0

4 years ago

Read 2 more answers

What is the surface area of the smaller prism?

Ymorist [56]

i just saw this class

the answer for the first one is 10m2

the answer for the second one is 90m2

have a nice day and be safe :)

5 0

4 years ago

The elevation of New Orleans, Louisiana is on 8 feet above sea level. The elevation of El Centro, California is 39 feet below se

Oxana [17]

Answer: 47

Step-by-step explanation:Find the absolute value of both which would be |8|+|-39|

Then you add 8 and 39 together which would be 47

4 0

3 years ago

Geometry Question attached below

Soloha48 [4]

I'm sure there's an easier way to do this, but this method does work:

First, AB = CD = CG + GF + FD, so FG = 2.

By the Pythagorean theorem, in triangle AFD we get

$1^2+3^2=A F^2\implies A F=\sqrt{10}$

and in triangle BCG,

$2^2+3^2=BG^2\implies BG=\sqrt{13}$

Angles AFD and EFG form a vertical pair, so they are congruent and have measure

$m\angle EFG=\tan^{-1}3$

Similarly, angles BGC and FGE are congruent and have measure

$m\angle FGE=\tan^{-1}\dfrac32$

Then the remaining angle in triangle EFG has measure

$m\angle FEG=\pi-\tan^{-1}3-\tan^{-1}\dfrac32$

We can solve for the lengths of FE and GE exactly by applying the law of sines:

$\dfrac{\sin\left(\pi-\tan^{-1}3-\tan^{-1}\frac 32\right)}2=\dfrac{\sin\left(\tan^{-1}3\right)}{GE}=\dfrac{\sin\left(\tan^{-1}\frac32\right)}{FE}$

$\implies GE=\dfrac{2\sqrt{13}}3,FE=\dfrac{2\sqrt{10}}3$

Let $s$ be the semiperimeter of triangle ABE, so that

$s=\dfrac{AB+BE+EA}2$

Then according to Heron's formula, the area of triangle ABE is

$\sqrt{s(s-AB)(s-BE)(s-EA)}=\dfrac{25}2$

5 0

4 years ago