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Guest Mar 29, 2020

#1**+1 **

First problem:

Call the point where the dashed line segment hits the base "X".

In triangle(BXA): (AX)^{2} + (XB)^{2} = (AB)^{2} ---> (AX)^{2} + (6)^{2} = (6.5)^{2}

---> A = 2.5

Drop a dashed line from C perpendicular to AB. Label the point where it intersects the base "Y",

Since BC = XY, AX + SY + YD = 2.5 + 9 + YD = 23 ---> YD = 11.5

In triangle(CDY): CY = 6 and YD = 11.5 ---> tan(D) = 6/11.5 ---> D = tan^{-1}(6/11.5)

---> D = 27.6^{o}

geno3141 Mar 29, 2020

#2**+1 **

Second problem:

Find AC:

Area( triangle(ABC) ) = 42 = ½·base·height = ½·BC·AC = ½·6·h

---> 42 = ½·6·h

---> 42 = 3h

---> h = 14

Since BC = 6 and BD = 22, DC = 22 - 6 = 16

In triangle(ADC): AC = 14 and DC = 16 ---> tan( D ) = 14/16 ---> D = tan^{-1}(14/16)

---> D = 41.2^{o}

geno3141 Mar 29, 2020

#4**0 **

Hi geno3141. I was working on my answer (just below) when you posted yours. It takes me longer because I explain why I did each step, as if the student were here and I was talking to him/her. I figure that since they need help with a relatively simple problem like this, they aren't familiar with the process. Do you think that my "talking" to the student is overdoing it? Is it too much? I've wondered this before but this is the first time I've asked. I'd appreciate your opinion, also from anybody else who would like to offer one. Thanks.

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Guest Mar 29, 2020

#5**0 **

I think it is a little too much detail...and too time consuming.. You are good to give explanations rather than just answers as so many do. I tend to believe that they ** should** have the knowledge of how to solve any problem they posted and know the basics of how to solve it given a little nudge/explanation in the right direction. I tend to give some direction and try to avoid giving a final answer, but let them do some of the calculations....o/w they will just run with the answer and not try to do anyhting themselves and not learn anything. I believe, given the number of repeated questions, a lot of the Q's are from homeschooled folks......they have to put something in the box on the computer screen to 'advance' or get the homework done and that becomes their objective: Fill in the box and move on....they may get 'advanced' or promoted, but then they are in over their heads at a level above their skills and more questions will be posted 'looking for answers' ..... YMMV

ElectricPavlov
Mar 29, 2020

#3**+1 **

Second problem:

First we'll solve for AC

The formula for the area of a triangle is A = (1/2)(base)(height)

The area of triangle ABC is 42

BC is 6 and we'll use it as the base

AC is the height of triangle ABC so 42 = (1/2)(6)(AC)

Do a little arithmetic and 42 = 3 • AC

Divide both sides by 3 and 14 = AC

Now we have learned that AC = 14

and because (22 – 6) = 16 CD = 16

We've got the hard part done, so now

it's time for a lilttle basic trigonometry

The tangent of angle ADC is

the opposite side over the adjacent side Tan(x) = AC/CD = 14/16

We could do the division by hand but

since we'll use the calculator to get the

arctan aka tan^{–1} let's just do the division

on the calculator too, for the convenience Tan(x) = 14/16 = 0**.**875

We see that the tangent is "close to" 1

so we know the angle will be "close to" 45^{o}

This is called "forecasting" or "guesstimating"

We use the arctan aka tan–1 function of

the calculator and find that the angle is Tan^{–1} (0**.**875) = 41**.**1859

ta da, and there you go (round it down a little) **angle x = 41.2 ^{o}**

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Guest Mar 29, 2020