Question

Help me please!!! These are going to be on my final.

Answer 1

Answer:

1. 70.32°

2. Point C to A is 28.50 miles.

3. 492.21 ft

4. 270.88 ft

5.45.77ft

6. Look at the last part of the explanation

Step-by-step explanation:

1. For this just remember that the longest side is opposite the biggest angle. So the largest angle will be the side facing 191ft.

$c^{2}=a^{2}+b^{2}-2ab(CosC)$

Refer to the picture to see how I referred each side.

a= 173 ft

b= 158 ft

c= 191 ft

C = unknown angle

To do this, we just plug in what we know and solve what we don't know.

$c^{2}=a^{2}+b^{2}-2ab(CosC)\\\\191^{2}=173^{2}+158^{2}-2(173)(158)Cos(C)\\\\36,481=29,929+24,964-54,668(Cos(C))\\\\36,481=54893-54668(Cos(C))\\\\36,481-54,893=-54,668(Cos(C))\\\\-18,412=-54,668(Cos(C))\\\\\dfrac{-18,412}{-54,668}=CosC\\\\0.3368=CosC\\\\Cos^{-1}0.3368=C\\\\70.32$

The biggest angle is 70.32°.

2. Here we use this same one. Please look at the attached figure to see the scenario. Because we are looking for b, we will use the b formula:

$b^{2}=a^{2}+c^{2}-2ac(CosB)$

Again we plug in what we know and solve what we don't know.

a=40 miles

b=?

c=50 miles

B=35°

$b^{2}=a^{2}+c^{2}-2ac(CosB)\\\\b^{2}=40^{2}+50^{2}-2(40)(50)(Cos35)\\\\b^2=1,600+2,500-4,000(Cos35)\\\\b^2=1,600+2,500-4,000(Cos35)b^2=4,100-3,287.61\\\\b^2=812.39\\\\\sqrt{b^{2}}=\sqrt{812.39}\\\\b=28.50$

Point C to A is 28.50 miles.

3. For this case we use the sine law:

$\dfrac{SinA}{a}=\dfrac{SinB}{b}=\dfrac{SinC}{c}$

What we are trying to look for is side b, which is the distance of point A to the kite.

If we draw the scenario, you can see that it makes a triangle. We have 2 known angles of the sides, and one unknown which is C. So how do we get this? The sum of all angles of a triangle is equal to 180. We know two sides, so all we need is the third side.

65 + 37 + C=180

102+ C = 180

C= 180-102

C = 78°

Now we know that C = 78°, we can use part of the formula:

Given would be:

B =37°

b = x

C=78°

c = 800ft

$\dfrac{SinB}{b}=\dfrac{SinC}{c}\\\\\dfrac{Sin37}{b}=\dfrac{Sin78}{800}\\\\\dfrac{Sin37}{b}=0.0012\\\\\dfrac{Sin37}{0.0012}=b\\\\492.21 ft=b$

4. We can get this using the trigonometric functions for a right triangle. Look at the figure to see your scenario.

What we are looking for is the opposite side, which is the side opposite of the given angle.

We are given the following:

h = 300ft

Θ = 62°

We are looking for the opposite, so we use this formula:

SOH

$Sin\theta=\dfrac{opposite}{hypotenuse}\\\\Sin62=\dfrac{opposite}{300ft}\\(Sin62)(300ft)=opposite\\\\264.88ft=opposite$

The height from the point where the string is held is 264.88ft.

But since it was held from the height of 6ft, we add this to our height.

264.88ft + 6ft= 270.88ft

5. Consider the illustration given to see the setup of the scenario. If we cut the diamond in half we create 2 triangles with angles of 90-45-45. So we have a known angle.

Using this information we can solve for the distance of the pitcher's mound to the third base using cosine law.

Given that:

c=65ft

a= 46ft

B= 45°

We can solve for c using the formula:

$b^{2}=a^{2}+c^{2}-2abCosB\\b^{2}=46^{2}+65^{2}-2(46)(65)Cos45\\b^{2}=2116+4225-5980(0.71)\\b^{2}=6341-4245.8\\\sqrt{b^{2}}=\sqrt{2095.2}\\b= 45.77ft$

6. Given the last attached image:

We can solve the rest of the triangle measures using cosine law:

$c^{2}=a^{2}+b^{2}-2abCosC\\c^{2}=8^{2}+11^{2}-2(8)(11)Cos(37)\\c^{2}=64+121-176(0.7986)\\c^{2}=185-140.5536\\\sqrt{c^{2}}=\sqrt{44.464}\\c = 6.67 units$

using sine law we can find the other angle:

$\dfrac{SinB}{b}=\dfrac{SinC}{c}\\\\\dfrac{SinB}{11}=\dfrac{Sin37}{6.67}\\\\\dfrac{SinB}{11}=0.09\\\\SinB = (0.09)(11)\\B = Sin^{-1}0.99\\B=81.89$

∠B= 81.89°

Lastly, we can solve for the last angle because all angles in a triangle sum up to 180°:

37° + 81.89°+x = 180°

118.89° + x = 180°

x = 180°-118.89°

x = 61.11°