All Common Core: High School - Geometry Resources
Example Questions
Example Question #5 : Understand That By Similarity, Side Ratios In Right Triangles Are Properties Of The Angles In The Triangle, Leading To Definitions Of Trigonometric Ratios For Acute Angles
Using the information from the triangle below, use the side ratios to find .
The definition of the sine of an angle is . So here we must determine which side is opposite of angle and which side is the hypotenuse of this triangle. We know that side is the hypotenuse since it is opposite of the right angle. Side is directly opposite of angle . So:
Example Question #6 : Understand That By Similarity, Side Ratios In Right Triangles Are Properties Of The Angles In The Triangle, Leading To Definitions Of Trigonometric Ratios For Acute Angles
Consider the following right triangle. Use trigonometric ratios to solve for sides and .
We are given angle and we need to find sides and . So first we need to think about what relation these sides are to angle . Side is the hypotenuse of the triangle since it is opposite of the right angle. So side must be the side adjacent to angle . Recall that the trigonometric ratio that corresponds to cosine is . We can solve for the missing sides by solving for the following equation.
It follows that:
Example Question #7 : Understand That By Similarity, Side Ratios In Right Triangles Are Properties Of The Angles In The Triangle, Leading To Definitions Of Trigonometric Ratios For Acute Angles
Using the information of the side lengths in the triangle below, use the side ratios to find the .
The definition of cosine of an acute angle is . So here we must determine what sides are adjacent and which is the hypotenuse. The hypotenuse is the easiest to pick out. This is the side that is directly across from the right angle in a right triangle. Our hypotenuse is . Now we must choose the adjacent side. Adjacent means next to, and since is our hypotenuse then must be our adjacent side. So:
Example Question #2 : Understand That By Similarity, Side Ratios In Right Triangles Are Properties Of The Angles In The Triangle, Leading To Definitions Of Trigonometric Ratios For Acute Angles
Find angle using trigonometric ratios.
60 degrees
50 degrees
75 degrees
45 degrees
45 degrees
We are given the adjacent side to angle and the hypotenuse of the triangle. We can use this to set up the trigonometric ratio which we know to be the definition of cosine. We can solve for angle using the following equation.
Example Question #9 : Understand That By Similarity, Side Ratios In Right Triangles Are Properties Of The Angles In The Triangle, Leading To Definitions Of Trigonometric Ratios For Acute Angles
Assume the two triangles below are similar. Using the fact that their side ratios are , what trigonomic function could this represent for angles and .
,
We must begin by manipulating our equation of the side ratios to get fractions that include both of the sides of the same triangles:
multiply both sides by
divide both sides by
If we look at angle we see that is the . Looking at angle we see that is also the . This is the definition of the tangent of an angle. So:
Example Question #10 : Understand That By Similarity, Side Ratios In Right Triangles Are Properties Of The Angles In The Triangle, Leading To Definitions Of Trigonometric Ratios For Acute Angles
Assume the following two triangles are similar. Using the fact that their side ratios are , what trigonomic function could this represent for angles and ?
,
,
,
,
,
We must begin by manipulating our equation of the side ratios to get fractions that include both of the sides of the same triangles:
If we look at angle , we see that is the . Looking at angle we see that is also the . This is the definition of the sine function. So:
Example Question #1 : Derive The Formula A = 1/2 Ab Sin(C) For The Area Of A Triangle By Drawing An Auxiliary Line From A Vertex Perpendicular To The Opposite Side
Is the following statement True or False?
We want to use the formula . Consider an obtuse triangle . We know the lengths of and , but only know the angle for . We are still able to use this formula.
False
True
True
There are two approaches to this problem. We are able to calculate the angle by using the Sine Law. The Sine Law states:
So we can set and solve accordingly for angle .
Our other option is to use the area formula we have been but altering it to correspond to angle . We would draw our vertical line down from the vertex as shown below and our formula would be in the form .
Example Question #1 : Derive The Formula A = 1/2 Ab Sin(C) For The Area Of A Triangle By Drawing An Auxiliary Line From A Vertex Perpendicular To The Opposite Side
Solve for x using the formula given that the area of the following triangle is (round to the second decimal place if needed).
Even though the formula is using sides and angle , this is a general formula and can be used with any angle in the triangle. Since we are now working with an obtuse angle rather than an acute angle, we need to do some more work to get the logic right.
Using the figure above, to be able to label the sides we are using correctly, we extend the original triangle horizontally past the obtuse angle and draw a vertical line down from the top vertex to form a right angle. This vertical line is . Angle for the supplementary (orange) triangle is . Using the fact that , we can set up our formula to be the following:
(either angle can be used and I will show this to be true)
(Using from the original triangle)
This shows that when using this formula for an obtuse angle, you can use either the supplementary angle you made, or the original. It is always helpful to draw this supplementary triangle in order to be able to visualize and understand logically how the formula is working for obtuse angles as well.
Example Question #2 : Derive The Formula A = 1/2 Ab Sin(C) For The Area Of A Triangle By Drawing An Auxiliary Line From A Vertex Perpendicular To The Opposite Side
Is the following statement True or False?
In order to use this formula, you only need to know two of the triangles sides' lengths.
True
False
False
In order to use this formula you need to either the height (which can be used to find the angle) or the angle (which can be used to find the height).
Example Question #3 : Derive The Formula A = 1/2 Ab Sin(C) For The Area Of A Triangle By Drawing An Auxiliary Line From A Vertex Perpendicular To The Opposite Side
Given the triangle below, what is the formula for finding the area?
The area of triangle is . Considering the entire triangle, . Solving for , . Therefore, .
The area of triangle is . Only considering the right triangle , . Solving for , . Therefore, .
The area of triangle is . Considering the right triangle , . Solving for , . Therefore, .
The area of triangle is . Considering the entire triangle, . Solving for , . Therefore, .
The area of triangle is . Only considering the right triangle , . Solving for , . Therefore, .
The formula for the area of a triangle is (base)(height). Since this is an obtuse triangle we need to break it into two right triangles by drawing the line down from the vertex perpendicular to the opposite side.
Now that we have two right triangles we can solve for the area of this triangle. Notice our base is and our height is . Plugging into the formula for the area of a triangle gives us . Most of the time, we will not have an exact length for , but we may have, or will be able to solve for, the lengths of and the angles . Using our relationship for right triangles, we know . In this case we will use as our angle. So
We can plug in for in our formula for area and we are left with .