Category: Geometry Chapter 11

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Geometry Ch11 test #7b

Question from Anya 
Course: Geometry
Chapter/Lesson/Problem11 test7b
Issue: I’m supposed to find Angle APM. To find the angle, I used tan APM=20/70, which gives 15.9 degrees. The solutions manual used sin APM=20/70, which gives 16.6 degrees. Sine is opposite/hypotenuse. However, it seems to me that the given measurements of the problem are opposite side (20) to the adjacent side (70). Which is correct? Thank you!
Answer by Dr.Callahan
Yes, you are correct. That should be tangent as you stated.
tan(APM) = 20/70
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Geometry Ch11.3 #38-41

Hannah asked:

I am doing Harold Jacobs Geometry,  I am on chapter 11 lesson 3.  I got to problems 39-41 and got stuck.  

Dr Callahan Answered:

Let us start at #38.

We can see that <ABC = 120 so <ADB = 30. Therefore we know AB = BD based on Theorem 10 page 159.

#39. 

AB = x  because given

BD = x  by substitution since AB  = BD

<BDC = 30 (since a right triangle with one angle 60)

2BC = BD (Them 51 on page 442)

2BC = x by substitution

BC = x/2 

#40

same logic using thrm 51 again  — of course starting from what we know from #39
CD = BC*SQRT(3)  Thrm 51 on page 442

BC = x/2 from #39

CD = x *SQRT(3) / 2

#41

We are to show how we get CD = 7/8 AB

AB = x

CD = x *SQRT(3) /2

We are to show CD/AB = 7/8 so we can divide what we have as AB and CD to get

CD/AB = (x *SQRT(3) /2 ) / x

CD/AB = SQRT(3) /2

Now you have to calculate SQRT(3) /2 to get about 0.866 and then do the math on 7/8 to get 0.875

They are not equal – but approximately equal.