Solve a Triangle

Question

Suppose that triangle ABC is a right triangle with a right angle at C and hypotenuse c. Also note that a is the length of the side opposite angle A and b is the length of the side opposite angle B.

Given that b = 1 and c = 30, determine the values indicated below. 

Round to four decimal places when needed. I found a already, I am having trouble getting the correct answer for A and B.

A is ? degrees

B is ? degrees

Answer 1

Draw the triangle and put on the known values.

A

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|___________________________________B

C

AC = 1 AB = 30

To find angle A, we know the side adjacent and the hypotenuse. These are in the definition of the Cos A.

Cos A = 1

30

So ask your calculator for the angle whose Cos is this value by using the Cos^-1 (1/30). Be sure your calculator is set in "degree" mode. If it is in "radian" mode you will get a wrong answer.

Cos^-1 (1/30) = 88.0898 degrees.

To find angle B, we know the side opposite and the hypotenuse. These are in the definition of the Sin B.

Sin B = 1

30

So ask your calculator for Sin-1 (1/30)

B = 1.9102 degrees.

You may verify your answers by adding the three angles for a total of 180 degrees.

90 + 88.0898 + 1.9102 = 180

Also, is the shortest side across from the smallest angle? Yes

Answer 2

sinB = b/c = 1/30; B sin^-1(1/30) = 1.9102°; B = 90° - A = 88.0898°

Answer 3

sin(B) = b/c = 1/30 since

b is the length of the side opposite angle B

and c is the length of the hypotenuse (opposite the right angle C)

so B = sin^–1(1/30) =

Calculate sin^–1(1/30) and round to four decimal places

The side opposite angle B is also

the side adjacent to angle A

So cos(A) = b/c = 1/30

which means that

A = cos^–1(1/30) =

Calculate cos^–1(1/30) and round to four decimal places

Note:

since 1/30 is a very small number

1. you can expect sin^–1(1/30) to be close to 0°

(It will be a little bit less that 2°)

1. you can expect cos^–1(1/30) to be close to 90°

(It will be a little bit greater that 88°)

Hope this helps. Add a comment if you have any questions.