Taylor's 7th(or 8th) Blog

The following is part two of my notes on lesson 9.1.

For this lesson we will only be using right triangles . We will be using Cos for these problems as well and only Cos for part 2 of this lesson.

To find Cos remember this:

Cos Theta = adjacent/ hypotenuse

(REMEMBER I DO NOT KNOW HOW TO PUT PICTURES ON THIS BLOG SO JUST DO AS I SAY AND REMEMBER THAT THE LENGTH OPPOSITE OF THE ANGLE IS THE LOWER CASE LETTER OF THAT ANGLE)

Triangle ABC is a right triangle. Use the following information to solve for it.

A=90 degrees a= 6
B= 65 degrees b=?
C= 25 degrees c=3

First, draw your triangle with the information that I gave you includes.

Cos 25= ?/6 Then, write out your equation like the formula I gave you.

b= 5.438 Finally, just multiply five on both sides, round to the third place after the decimal and you get this as your answer.

Triangle ABC is a right triangle. Use the following information to solve for it.

A=90 degrees a= 10
B= ? degrees b=6
C= ? degrees c=7

First, draw your triangle with the information that I gave you includes.

Cos C= 6/10 Then, write out your equation like the formula I gave you.

C= Cos^-1 (6/10) Then, you need to find the inverse.

C= 53. 130 degrees Then, solve and round to the third place after the decimal. Then, you would normally need to make sue that there is no other possible angle but since the positive form of
Cos is on the first and forth quadrant, you cannot do it for the forth quadrant is up to 360 degrees and triangles can only go up to 180 degrees so this is your only answer.

180 degrees -90 degrees -53. 130 degrees= 36.87 degrees Finally, you just have to subtract 90 degrees and 53. 130 degrees from 180 degrees and you will get 36.87 degrees for angle B.

So your answers will be C= 53. 130 degrees and B= 36.87 degrees.

THAT IS ALL FOR THIS PART OF MY LESSON 9.1 PART 2 NOTES !! SO UNTIL NEXT TIME JA NE!!!! (ja ne is goodbye in Japanese)