Trigonometry : Triangles
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Example Questions
Example Question #1 : Trigonometric Applications
Given a right triangle where 
Since the triangle in question is a right triangle we can use the Pythagorean Theorem. First, we must determine which sides we are given. Since the function we are given is sine, we know that we are given the opposite side and the hypotenuse. Therefore, setting up the equation:
Where, 
Solving the above equation:
We toss out the negative solution since the length of a side must be positive.
Example Question #2 : Trigonometric Applications
Given a right triangle where 
Since the triangle in question is a right triangle we can use the Pythagorean Theorem. First, we must determine which sides we are give. Since the function we are given is tangent, we know that we are given the opposite and adjacent sides. Therefore, setting up the equation:
Where, 
Solving the above equation:
We toss out the negative solution since the length of a side must be positive.
Example Question #1 : Trigonometric Applications
Given a right triangle where 
Since the triangle in question is a right triangle we can use the Pythagorean Theorem. First, we must determine which sides we are given. Since the function we are given is cosine, we know that we are given the adjacent side and hypotenuse. Therefore, setting up the equation:
Where, 
Solving the above equation:
We toss out the negative solution since the length of a side must be positive.
Example Question #4 : Trigonometric Applications
Given the accompanying triangle where 
We are given the opposite side, with respect to the angle, along with the angle. Therefore, we utilize the sine function to determine the length of the hypotenuse:
Substituting the given values:
Cross multiplying:
Solving for 
Example Question #1 : Triangles
Given the accompanying right triangle where 
We are given two sides of the right triangle, namely the hypotenuse and the opposite side of the angle. Therefore, we simply use the sine function to determine the angle:
In order to isolate the angle we must apply the inverse sine function to both sides:
Example Question #6 : Trigonometric Applications
Given the accompanying right triangle where 
We are given two sides of the right triangle, namely the hypotenuse and the adjacent side of the angle. Therefore, we simply use the cosine function to determine the angle:
In order to isolate the angle we must apply the inverse cosine function to both sides:
Example Question #5 : Trigonometric Applications
All the squares are equal, and there are 6 squares in the figure.
What is the value of 
Example Question #1 : Right Triangles
Find the value of 
The first two things to recognize regarding our tirangle are 1) it is a right triangle and 2) it is an isosceles triangle. The two congruent sides tell us that the two non-right angles are also congruent, and a little quick math tells us that they each equal 45 degrees. This means our right triangle is not just any right triangle but a 45-45-90 triangle.
This is important because the sides of every 45-45-90 triangle follow the same ratio. The two legs are obviously always congruent to each other (being isosceles), but to find the hypotenuse, we simply have to multiply the length of a leg by 
Given this fact we would be in good shape if we had the length of a leg and needed the hypotenuse. But we have the hypotenuse and need the leg, which we means we need to work backwards going this way, we need to divide the length of the hypotenuse by 
However, general practice in mathematics doesn't allow us to leave a square root in the denominator. We solve this problem by rationalizing the denominator, which is accomplished by multiplying the numerator and the denominator by 
This effectively eliminates the square root in the denominator and provides our answer.
Example Question #2 : 45 45 90 Triangles
A triangle has three angles 
A triangle with the sum of two angles equaling the third is a 
Example Question #3 : Triangles
The following figure was made by beginning with a square. The midpoints of the four sides of the square were then joined to form another square. The process was repeated to form a third square and finally once more to form the fourth and smallest square in the middle, which has a side length of 
We begin by realizing that the midpoints of the sides of our outer square divide each side in half. Furthermore, the sides of our second square connecting these midpoints form four right triangles in each corner of our largest square.
But these right triangles are special right triangles. They are 45-45-90 triangles, which means we can find the hypotenuse (and thus the side of our second square) by multiplying the length of the leg by 
We now repeat the process, beginning by forming four new 45-45-90 triangles
To find the hypotenuse of each of these triangles (and thus the side length of our third square), we simply multiply by 
We then repeat the process one final time, multiplying by 
Our final hypotenuse and thus the side of our innermost square is 
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