Trigonometry : Triangles

Study concepts, example questions & explanations for Trigonometry

varsity tutors app store varsity tutors android store

Example Questions

← Previous 1 3 4 5 6 7 8 9 16 17

Example Question #1 : Trigonometric Applications

Given a right triangle where , find the missing side.

Possible Answers:

Correct answer:

Explanation:

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,  and  are given.

Solving the above equation:

We toss out the negative solution since the length of a side must be positive.

Example Question #2 : Right Triangles

Given a right triangle where , find the missing side.

Possible Answers:

Correct answer:

Explanation:

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,  and  are given.

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 , find the missing side.

Possible Answers:

Correct answer:

Explanation:

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,  and  are given.

Solving the above equation:

We toss out the negative solution since the length of a side must be positive.

Example Question #4 : Right Triangles

Given the accompanying triangle where  and , determine the length of the hypotenuse.

Right_triangle

Possible Answers:

Correct answer:

Explanation:

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 #3 : Trigonometric Applications

Given the accompanying right triangle where  and , determine the measure of  to the nearest degree.

Right_triangle

Possible Answers:

Correct answer:

Explanation:

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 : Right Triangles

Given the accompanying right triangle where  and , determine the measure of  to the nearest degree.

Right_triangle

Possible Answers:

Correct answer:

Explanation:

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 #7 : Right Triangles

Question 

All the squares are equal, and there are 6 squares in the figure.

What is the value of   ?

Possible Answers:

Correct answer:

Explanation:

Solution

Example Question #1 : 45 45 90 Triangles

Find the value of  in the triangle below.

16

Possible Answers:

Correct answer:

Explanation:

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 . Therefore,

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 : Right Triangles

A triangle has three angles such that  and  together are as much as . What is the ratio of the longest side to the shortest?

Possible Answers:

Correct answer:

Explanation:

A triangle with the sum of two angles equaling the third is a  triangle

Example Question #10 : Right 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 .  Find the value of .

18

Possible Answers:

Correct answer:

Explanation:

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.

18

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 .  Therefore the length of a side of our second square is .

18

We now repeat the process, beginning by forming four new 45-45-90 triangles 

18

To find the hypotenuse of each of these triangles (and thus the side length of our third square), we simply multiply by  again.

18

We then repeat the process one final time, multiplying by  again.

18

Our final hypotenuse and thus the side of our innermost square is .

← Previous 1 3 4 5 6 7 8 9 16 17
Learning Tools by Varsity Tutors