4.5 Applications of Right Triangle Trigonometry

 

When one knows one of the acute angles of a right triangle and the length of one of the sides, one can solve for the length of the other two sides using trigonometric functions of the given angle.

 

It is a convention that if a vertex of a triangle is denoted by an upper case letter, then the side opposite that vertex is denoted by the corresponding lower case letter.  Thus the side opposite vertex A is a, the side opposite B is b, etc.

 

In the following examples, it will always be assumed that C is the right angle.

 

 

 

Example:  Let A = 30o, and a = 10.  Find b and c.

 

Form a ratio between the unknown side and the known side, with the known side in the denominator.  Determine which trigonometric function of the known angle represents that ratio, then equate the two.

 

To find c, we solve c / 10 = csc 30o.  Thus, c = 10 csc 30o = 10 ( 2 ) = 20.

 

To find b, we solve b / 10 = sec 30o.  Thus, b = 10 sec 30o = 10 (   ) = .

 

Exercise 4.5.1

 

Let B = 45 o, and c = 12.  Find a and b.

 

Solution

 

Exercise 4.5.2

 

Let A = 39.7 o, and a = 3.46.  Find b and c.  Use a calculator.  There is no cotangent function on a calculator.  You will have to use 1 / tangent instead.  Likewise, you will have to use 1 / sine instead of cosecant since there is no cosecant function on a calculator.

 

Solution

 

Exercise 4.5.3

 

A flagpole stands in the middle of a flat, level field.  Fifty feet away from its base a surveyor measures the angle to the top  of the flagpole as 48°.  How tall is the flagpole?

 

Solution

 

Exercise 4.5.4

 

A 100 foot wharf sits along the bank of a river.  A surveyor stands directly across the river from one end of the wharf.  From where he stands the angle between the lines of sight to the two ends of the wharf is 31°.  How wide is the river?

 

Solution

 

Exercise 4.5.5

 

Standing across the street 50 feet from a building, the angle to the top of the building is 40°.  An antenna sits on the front edge of the roof of the building.  The angle to the top of the antenna is 52°.  How tall is the building.  How tall is the antenna itself, not including the height of the building?

 

Solution

 

Exercise 4.5.6

 

Standing on one bank of a river, an explorer measures the the angle to the top of a tree on the opposite bank to be 27°.  He backs up 50 feet and re-measures the angle to the top of the tree at 22°.  How wide is the river?

 

Solution

 

The Pythagorean Theorem and the Right Triangle Method

 

The famous Pythagorean Theorem of antiquity states that the sum of the areas of the two squares constructed on the sides of a right triangle equals the area of the square constructed on the hypotenuse.  The hypotenuse is the side opposite the right angle.

 

Applied to the right triangle depicted above, this principle can be expressed by the algebraic equation:

 

a2 + b2 = c2

 

Given the value of one of the six trigonometric functions of an angle A and given the quadrant in which the terminal side of A lies when A is in standard position, the Right Triangle Method allows one to compute the values of the other five trigonometric functions of A.  All that is needed is the basic definition of the six trigonometric functions, where we used x, y and r instead of a, b and c.  That is x2 + y2 = r2.

 

Example:  Given that angle A lies in quadrant II and sin A = 2 / 3, draw and correctly label the sides of a right triangle in such a way that the find the other five trigonometric functions of A may be found using the basic definitions.

 

Answer:  Draw a right triangle with base angle A.  Place a value of 2 on side a and a value of 3 on the hypotenuse.  Using the Pythagorean Theorem, we write x2 + 22 = 32 and conclude that x2 = 5.  So either x =  or x = .  But since A lies in quadrant II and since x coordinates in quadrant II are negative, it must be the case that x = .  So the triangle must be labeled as follow:

 

 

 

 

Now that all three sides are properly labeled, we can compute the other five trigonometric functions directly from the definitions.

 

Example:  Tan A = 3 / 4 and A is in quadrant III.  Notice, that in quadrant II both x and y must be negative. [By the way, r will never be negative.]

 

Answer:

 

Exercise 4.5.7

 

Given that A is in quadrant IV and sec A = 4, correctly label the right triangle and use it to find the values of the other five trigonometric functions of A.

 

Solution

 

Exercise 4.5.8

 

Given that A is in quadrant II and cot A = t < 0, find the other five trigonometric functions of A as functions of t.

 

Solution

 

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