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3 years ago

Two isosceles triangles are shown above. If b = 180-4a and a=35, What is the value of c?

Mathematics

1 answer:

Ymorist [56]3 years ago

4 0

Answer:

a=35 given

b=40

c=110

I couldn't complete see c. Please look at the picture to see what I assumed it to be.

Step-by-step explanation:

Hmmm... I guess a and b are not vertical.

We are given b=180-4a and a=35 so b=180-4(35)=180-140=40.

So b=40.

Isosceles triangles always have congruent base angles. So let's call both of the base angles in the bottom triangle x.

That means x+x+40=180.

We need to solve this for x.

Combine like terms:

2x+40=180

Subtract 40 on both sides:

2x=140

Divide both sides by 2:

x=140/2

Simplify:

x=70

So I'm assuming that c and it's adjacent angle are sitting on a straightedge together which means 70+c=180.

70+c=180

Subtract 70 on both sides:

c=180-70

Simplify:

c=110

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You are a lifeguard and spot a drowning child 60 meters along the shore and 40 meters from the shore to the child. You run along

sukhopar [10]

Answer:

The lifeguard should run across the shore a distance of 48.074 m before jumpng into the water in order to minimize the time to reach the child.

Step-by-step explanation:

This is a problem of optimization.

We have to minimize the time it takes for the lifeguard to reach the child.

The time can be calculated by dividing the distance by the speed for each section.

The distance in the shore and in the water depends on when the lifeguard gets in the water. We use the variable x to model this, as seen in the picture attached.

Then, the distance in the shore is d_b=x and the distance swimming can be calculated using the Pithagorean theorem:

$d_s^2=(60-x)^2+40^2=60^2-120x+x^2+40^2=x^2-120x+5200\\\\d_s=\sqrt{x^2-120x+5200}$

Then, the time (speed divided by distance) is:

$t=d_b/v_b+d_s/v_s\\\\t=x/4+\sqrt{x^2-120x+5200}/1.1$

To optimize this function we have to derive and equal to zero:

$\dfrac{dt}{dx}=\dfrac{1}{4}+\dfrac{1}{1.1}(\dfrac{1}{2})\dfrac{2x-120}{\sqrt{x^2-120x+5200}} \\\\\\\dfrac{dt}{dx}=\dfrac{1}{4} +\dfrac{1}{1.1} \dfrac{x-60}{\sqrt{x^2-120x+5200}} =0\\\\\\ \dfrac{x-60}{\sqrt{x^2-120x+5200}} =\dfrac{1.1}{4}=\dfrac{2}{7}\\\\\\ x-60=\dfrac{2}{7}\sqrt{x^2-120x+5200}\\\\\\(x-60)^2=\dfrac{2^2}{7^2}(x^2-120x+5200)\\\\\\(x-60)^2=\dfrac{4}{49}[(x-60)^2+40^2]\\\\\\(1-4/49)(x-60)^2=4*40^2/49=6400/49\\\\(45/49)(x-60)^2=6400/49\\\\45(x-60)^2=6400\\\\$

$x$

As $d_b=x$ , the lifeguard should run across the shore a distance of 48.074 m before jumpng into the water in order to minimize the time to reach the child.