Finding the angles of a rectangle when there's a diagonal

[Andreaaiii]

A rectangle has length 25 mm and width 18 mm. Find the angles between the diagonals of the rectangle.

In my math book, the answer is: 71.5 degrees and 108.5 degrees. I'm wondering how do you solve this?

Thank you.

 

[stapel]

A rectangle has length 25 mm and width 18 mm. Find the angles between the diagonals of the rectangle.

In my math book, the answer is: 71.5 degrees and 108.5 degrees. I'm wondering how do you solve this?

Draw the rectangle, with the longer sides horizontal and the shorter sides vertical (so we're looking at the same picture). Label its sides with the given dimensions. Draw one of its diagonals; say, the one that goes from the lower left-hand corner to the upper right-hand corner.

You now have a right triangle, formed by the lower horizontal line with length 25, the right-hand vertical line with length 18, and the hypotenuse formed by the diagonal. Use the Pythagorean Theorem, if you like, to find the length of the hypotenuse.

Label the angle at the lower left-hand vertex as \(\displaystyle \alpha .\) You can use a trig ratio to find the measure of this angle. For instance, you could use \(\displaystyle \cos(\alpha)\, =\, 18/25,\) and then use \(\displaystyle \alpha \,=\, \cos^{-1}(18/25)\) to find the measure of that base angle.

You know that the diagonals cross at the center of the rectangle. Draw in the other diagonal, and then drop a vertical line from that center crossing point down to the base. The second diagonal (counting only the part from the center down to the lower right-hand corner of the original rectangle) creates an isosceles triangle. You know the measure of the one base angle (from the right triangle you were working with earlier), so you therefore know the measure of the other base angle. Since you know the angle-sum of any triangle, you can then find the measure of the angle at the "top" vertex of the isosceles triangle. This is also one of the two angles you need, in order to answer the original question.

Since the angles forming a straight line must sum to 180°, then you can find the measure of the other vertical angle formed by the diagonals.

If you get stuck, please reply showing your work in following the step-by-step instructions above. Thank you!

 

[Steven G]

Label the angle at the lower left-hand vertex as \(\displaystyle \alpha .\) You can use a trig ratio to find the measure of this angle. For instance, you could use \(\displaystyle \cos(\alpha)\, =\, 18/25,\) and then use \(\displaystyle \alpha \,=\, \cos^{-1}(18/25)\) to find the measure of that base angle.you can then find the measure of the angle at the "top" vertex of the isosceles triangle. This is also one of the two angles you need, in order to answer the original question.

I am not sure if you wanted to make the trig mistake on purpose so not to give everything away to the OP but it turns out that \(\displaystyle \tan(\alpha)\, =\, 18/25,\) and then ....

 

[pka]

Draw the rectangle, with the longer sides horizontal and the shorter sides vertical (so we're looking at the same picture). Label its sides with the given dimensions. Draw one of its diagonals; say, the one that goes from the lower left-hand corner to the upper right-hand corner.
Label the angle at the lower left-hand vertex as \(\displaystyle \alpha .\) You can use a trig ratio to find the measure of this angle. For instance, you could use \(\displaystyle \cos(\alpha)\, =\, 18/25,\) and then use \(\displaystyle \alpha \,=\, \cos^{-1}(18/25)\) to find the measure of that base angle.

I am not sure if I follow the construction.

If that is correct then \(\displaystyle \alpha = \arctan \left( {\frac{{25}}{{18}}} \right)\)

By the external angle theorem \(\displaystyle m(\beta ) = 2m(\alpha )\)
see here

 

[Andreaaiii]

Thank you!

Thank you all for the feedback. I have a better understanding. The diagram really helped me visualize the rectangle. I only worked with one diagonal, I didn't know I needed two.

Also, when I solved it. Why do you have to have 2*arc tan (18/25)? What is the purpose of multiplying it by 2?

Thank you.

 

[Andreaaiii]

Thank you

I am not sure if I follow the construction.
View attachment 4839
If that is correct then \(\displaystyle \alpha = \arctan \left( {\frac{{25}}{{18}}} \right)\)

By the external angle theorem \(\displaystyle m(\beta ) = 2m(\alpha )\)
see here

Thank you very much for the diagram. What is the purpose of the "external angle theorem" ?

 

[pka]

Thank you very much for the diagram. What is the purpose of the "external angle theorem" ?

The measure of an exterior angle of a triangle equal to the sum of measures of the two opposite interior angles. In this case we have an isosceles triangle so \(\displaystyle
m(\beta)=2m(\alpha).\)

 

[stapel]

I am not sure if you wanted to make the trig mistake on purpose so not to give everything away to the OP but it turns out that \(\displaystyle \tan(\alpha)\, =\, 18/25,\) and then ....

Thank you!