Do you have an image or something? Because now it's difficult to know how you've named the triangle and which side is which side ... For example what is 'MN'?
I have a right angled triangle and I am given the following information;
I have two angles, one is the normal 90 degree angle and the other angle is 65 degrees in the opposite vertex.
I am told that the right angled triangle has a parallel line to BC.
To imagine the triangle the 90 degrees corner is on the lower left and the 65 degrees is on the lower right vertex. The parallel line is 5 cm long and I am asked to find the length of the side AB, which is the adjacent side of a right angled triangle.
I have used the tan rule;
tan 65 = AB / MN
AB / MN = tan 65
tan 65 = AB / MN = 2.145
muliplied both sides by 5
AB = 2.145 x 5
AB = 10.7 cm (1dp)
My confusion is whether I used the right sin rule as the cosine rule would give an answer of 11.8 cm
They are both close to each other in their solutions, could somebody please advise how I determine the right cosine rule to use when working with parallel lines in triangles please.
I knew when I was posting the thread that defining the right angled triangle would be difficult so I will do my best here;
Draw a vertical line AB, A is at the top. Draw a horizontal line 5cm to the right. Draw the line of the hypotenuse.
The triangle is A at the top of the line known as the opposite, B is at the bottom and C to at the right hand side.
You should end up with a right angled triangle with the 90 degrees in the left lower corner.
Now Midpoint in the triangle draw another horizontal line, this is parallel to the horizontal line at the bottom, which is referred to as adjacent. This line is MN
The angles;
angle theta is 65 degrees, this is opposite the right angle of 90 degrees.
When you look at the triangle with the parallel line drawn there is one triangle split in two.
I hope you can follow my logic?
It's called the 3, 4, 5 triangle, with the right angle in the lower left corner. The line MN is drawn midpoint through the centre of the triangle, so in effect you get to parallel lines which can bot be thought of as opposite side(s).
Hoe this makes it clearer?
If I understand the problem correct then your calculations in your first post aren't entirely correct because along your calculations the line MN and AC are overlapping which is not true by the given. Do you understand what I mean? ...
What you've calculated is the side AM.
I'll try to explain it without a design. Ok if you have your design you'll see the line MN divides the triangle ABC in two pieces, the upper piece is a right angle triangle and the lower piece is a (right angle) trapezium. Draw a line out of the point N (which lies on the hypotenuse) perpendicular to the line BC so you divides the trapezium in a right angle triangle and a square or rectangle? What do you see on the given picture: a square or a rectangle? ...
Well, divide this right angle trapezium in two pieces: a right angle triangle and another piece which has to be a square or a rectangle? But because I don't have the design and you do I can't see if it's a rectangle or a square ...
As previously I have drawn the triangle and extended the line MN to the left, then extended the line BC to the left and then joined them up with a line as the hypotenue is placed with a right angle triangle, the result is a trapezium with one pair of parallel lines. The top parallel line is MN at 5cm in length, the bottom parallel line is BC, which I have no length of.
So the data I have now got for this trapezium is MN 5cm, two lower corner angles of 65 degrees, and two top corner angles of 115 degrees.
Not sure what to do next, even if I divide by two, I am still lost?
OK sorry didn't mean to upset any members, I should have created a picture to illustrate my requests. It should be appreciated that not all computer users are experts and are learning as we go along.
The question I have is here re-stated;
To find the length of the side AB, is this cos 65 x 5 = 2.11cm and then double it for the length MB so = 2.11 x 2 = AB 4.22 cm, or
tan 65 x 5 = 10.7 cm and then double it for the length AB so = 10.7 x 2 = 21.4 cm.
Any help much appreciated
Thanks
David
I think I am probably looking here at a tangent ratio and the triangles are similar.
I am thinking that;
MN / AM = BC / AB, but then I have two unknowns?
5 / 10.7 = x / x
Similar if this problem is a ratio problem, then;
tan theta = BC / AB, and I still have two unknowns?
I have worked out the top section of the triangle and concluded the following results;
The triangle is; AMN
AM = 10.7
MN = 5
AN = 11.81
Is there a method that allows me to use these results to find the length of the sides of the triangle ABC?
Thanks
David
OK I have solved the problem now.
In the triangle previously, it is a right angle 3,4,5 type. The confusion was the parallel line MN at 5cm, I didn't know what to do with it.
I worked out that AM = 10.7cm, AN = 11.81cm from the given data.
Using similarity;
BC / MN = AB / AM = AC / AN
AM = 10.7, MB = 10.7, so AB = 10.7 + 10.7 = 21.4
21.4 / 10.7 = x / 5
21.4 x 5 / 10.7 = 10
Length BC = 10cm
Tan 65 x 10 = 21.4
AB = 21.4 cm
David