Happy Mr. Power meant a demanding lesson
Mr. John Power looked very happy again. Many students started thinking that he must have something up his sleeve. Last class, came with the same happy face, he had asked class to determine the value of pi using trigonometry. They all wondered what had he come up with this time.
Mr. Power: Today we are going to learn about right angle, obtuse and acute triangles. Tinku, what is a right angle triangle ?
Tinku was amused at how easy this question was and blurted out: A right angle is a triangle in which one of the angles is a right angle that is 90°.
Mr. Power: Who can tell me what are obtuse and acute triangles ?
Several students raised their hands but Mr. Power asked Tracy.
Tracy: In an obtuse triangle – one of the angles is larger than 90° but in an acute triangles all three angles are less than 90°.
Classifying triangles based on cosines
Mr. Power: Sara, please write the trigonometric functions of the three type of triangle.
Sara went to the board and made this table for the largest angle (BAC) in each type of triangle (see Figure).
|Sin BAC||>0 but < 1||1||>0 but < 1|
|Cos BAC||<0 (negative)||0||>0 but < 1|
|Tan BAC||<0 (negative)||∞||>0 but < ∞|
Mr. Power: Good work Sara. Now if you had to define the three types of triangles based on trigonometric relationships, what function will you choose ?
Sara: Cos, because it is simple: negative for obtuse, 0 for right angle and positive for acute angle triangle.
Many students were awed how one could use trigonometry to define the three types of triangles.
Mr. Power was not done yet. He asked if anyone remembered the Pythagoras Theorem they had learned in Geometry. Many students raised their hands and he called on a student named Tracy to describe the theorem.
Tracy: In a right angle triangle the square of the length of the hypotenuse equals sum of the squares of the lengths of the other two sides.
Mr. Power: If the length of the hypotenuse is c and the other sides are of lengths a and b, would you say then c2 = a2 + b2 ?
Tracy: Yes, Mr. Power.
Does c2 = a2 + b2 work for obtuse triangles
Mr. Power: Alright class. I want you to sit in groups of three and figure out if this relationship will also apply to obtuse angle triangles with the length of the longest side being c and the other two sides being a and b. I will give you a hint. In the triangles ABC on the board, draw a vertical line from the corner C onto the base AB.
After about 10 minutes, Tommy’s group said that they had figured out that for an obtuse triangle c2 > a2 + b2. He came to the board and wrote this.
For an obtuse triangle ABC, the length of the side BC which is opposite to the obtuse angle being c, the base AB being a and side AC is b. We drew a vertical line CD which met the extended base AB at D. Let the length of BD be h and AD be x.
Then from the large right angle triangle DCB, h2 = c2 – (a+x)2.
From the triangle DAC, h2 = b2 – x2.
Therefore c2 – (a+x)2 = b2 -x2.
Adding (a+x)2 to both sides of the equation, we get c2 = (a+x)2 + b2 -x2 which is
c2 = a2+2ax +x2 + b2 -x2 or c2 = a2 + b2 +2ax.
Because a, b, c and x are all positive c2 > a2 + b2.
Mr. Power along with most of the students accepted the proof and were impressed.
Describing triangle types with algebraic equations
Mr. Power: Well done Tommy and your partners Sara and Johnny for showing that in an obtuse triangle c2 > a2 + b2.
Remember Tracy said that for a right angle triangle c2 = a2 + b2. I am going to add that for an acute angle triangle with the longest side being of length c, c2 < a2 + b2 and your homework is to prove it.
For the obtuse, right angle and acute triangles cosine of the angle opposite to the longest sides will be negative, 0 and positive, respectively. I am going to add that we could write that c2 > a2 + b2, c2 = a2 + b2 and c2 < a2 + b2, respectively. Here is the summary of today’s lesson.
|Cos BAC||<0 (negative)||0||>0 but < 1|
|Value of the longest side c||c2 > a2 + b2||c2 = a2 + b2||c2 < a2 + b2|
Prove that for an acute angle triangle with sides of length a, b and c and the longest side being of length c, c2 < a2 + b2.
Given: An acute angle triangle ABC, side BC is the longest side of length c and length of side AB = a and AC = b.
Draw a vertical line BD which meets the base at D. Let the length of BD be h and BD be x, then AD of length = a-x.
From the right angle triangle BCD, c2 = h2 + x2
From the right angle triangle ACD, b2 = h2 + (a-x)2 or h2 = b2 – (a-x)2
Therefore c2 = b2 – (a-x)2 + x2 or c2 = b2 + x2 – a2 – x2 +2ax
x2 or c2 = b2 – a2 +2ax or c2 = b2 + a2 – 2a2 +2ax or
c2 = b2 + a2 – 2a (a – x)
Because a is larger than x, a-x are positive, and hence
c2 < (a2 + b2).