Question

Help me please these going to be on my final.

Answer 1

7. 46.39 miles

8. 109 cm

9. 18 cm

Explanation:

7. Look at the attached picture so you can picture the scenario.

The distance from City A to City B is 85 miles North. Notice that the plane only got to travel 20 miles North because of a cloudbank and redirected 108 degrees NORTH OF EAST, 20 miles towards that direction.

The new path makes a triangle, as shown in the picture.

The angle of the triangle was solved this way.

108° - 90° = 18°

The length of a was solved by getting how many miles the plane didn't travel in its original path.

85 miles - 20 miles = 65 miles.

So now we have our new given:

a = 65 miles

b = 20 miles

c = ?

C = 18°

Using the cosine law we can solve for the c, which is the distance of the plane from City B.

`c^(2)=a^(2)+b^(2)-2abCosC\\\\c^(2)=65^(2)+20^(2)-2(65)(20)Cos18\\\\c^(2)=4225+400-2600(Cos18)\\\\c^(2)=4625-2472.75\\\\\sqrt{c^(2)}=√(2152.25)\\\\c=46.39miles`

8. Again look at the attached. So as you can see we make another triangle. We can solve this with the cosine law once again. Just remember that the longer diagonal is opposite the biggest angle.

Given that:

a = 54cm

b = 78 cm

C = 110°

`c^(2)=a^(2)+b^(2)-2abCosC\\\\c^(2)=54^(2)+78^(2)-2(54)(78)Cos110\\\\c^(2)=2916+6084-8424(Cos110)\\\\c^(2)=9000-(-2281.18)\\\\\sqrt{c^(2)}=√(11881.18)\\\\c=109cm`

9. Again, the shorter diagonal is facing the smaller angle. But we use the same cosine law to find it, but we use the formula of a, since a is what we are missing.

Given that:

a=?

A = 58°

b = 21cm

c= 15cm

`a^(2)=b^(2)+c^(2)-2bc(CosA)\\\\a^(2)=21^(2)+15^(2)-2(21)(15)(Cos58)\\\\a^(2)=441+225-630(Cos58)\\\\a^(2)=666-333.85\\\\\sqrt{a^(2)}=√(332.15)\\\\a = 18cm`