let a (0,0), b(7,0), and c(7,5). point d is located so that angle acd is a right angle and the tangent of angle dac is 5/7. find coordinates for d.

Answer 1

There may be a slightly easier way to get this answer.

Draw a figure.

As Emily W. observed there are two possible answers.

Let d be the point northwest of c and e (the other possibility for d) be southeast of c.

Angle acd and angle acb are equal since both have tangent=5/7.

Line dce has equation -7/5=(y-5)/(x-7) and line ad has equation y=35x/12 since angle dab is twice angle cab.

The intersection of these 2 equations gives the co-ordinates of d.

The 2 triangles are congruent so that e must be on the x-axis and the y value is 74/7.

Answer 2

Follow along with your drawing.

AB = horizontal line 7 units long

BC = vertical line 5 units tall

AC = hypotenuse of original right triangle

D needs to be placed somewhere such that ACD is a right angle and the tangent of DAC is 5/7

If C is the location of the right angle, we need to find the perpendicular line to AC that passes through C and will contain the point D.

Slope of AC = 5/7

Slope of DC = negative reciprocal = -7/5

Now we need to make an equation for this line. We know the slope of the line is -7/5 and we know it just contain the point C (7,5). We can use point slope form

y - y1 = m(x-x1)

y - 5 = -7/5(x - 7) EQUATION 1

This is one of our key equations to solve this problem

If we follow this line, we can see that there are 2 possible places to put D: up and to the left of C, and down and to the right of C, that would make a right triangle. There will be 2 answers.

Next, we know that the tangent of DAC = 5/7

tangent = opposite/adjacent

Since angle C is the right angle, the side across from it, AD, is the hypotenuse of our new triangle.

The opposite side is DC

The adjacent side is AC

We can find the value of AC because it is the hypotenuse of our original triangle ABC. Let’s use Pythagorean theorem:

a^2+b^2 = c^2

7^2 + 5^2 = c^2

49+25 = 74=c^2

c = square root(74)

^^ this is the adjacent side of our new triangle

We can find out how long our opposite side needs to be by setting it equal to our tangent

tangent = 5/7 = opp/adj

5/7 = opp/ sqrt(74)

multiply both sides by sqrt(74)

Opposite side = 5/7 * sqrt(74)

This means the DISTANCE between C and D must be 5/7*sqrt(74)

Let’s set up the distance formula to find a relationship between C and D

distance formula

(x2-x1)^2 + (y2-y1)^2 = distance^2

We know the coordinates of C are (7,5) so let’s use x1= 7 and y1 = 5

Then we can say that the other x and y are the coordinates of D that we are missing

(x - 7)^2 + (y - 5)^2 = distance^2

our distance is what we just found —> the length of the opposite side, 5/7*sqrt(74)

distance^2 = 5^2 / 7^2 * [sqrt(74)]^2

(a square and a square root cancel each other out to just become 74)

distance ^2 = 25/49 * 74 = 1850/49

put into our distance equation

(x-7)^2 + (y-5)^2 = 1850/49 EQUATION 2

Our 2 equations look pretty similar

(y - 5) = -7/5(x - 7)

(x - 7)^2 + (y - 5)^2 = 1850/49

We want to make the top equation look as much like the bottom one as we can so we can combine them together. The top equation is missing some squares, so let’s square both sides.

(y - 5)^2 = [-7/5(x-7)]^2

(y - 5)^2 = 49/25 (x - 7)^2

Looks like we have an expression for (y - 5)^2 that we can SUBSTITUTE into the second equation!

(x - 7)^2 + (y - 5)^2 = 1850/49 Substitute for (y-5)^2

(x - 7)^2 + 49/25(x - 7)^2 = 1850/49

We have two parentheses that are (x - 7)^2 but it would be painful to foil this out. Instead, I am going to replace it with something simpler.

big X = x-7 now I will rewrite my equation

X^2 + 49/25X^2 = 1850/49

This is much easier to solve. Combine like terms

X^2 + 49/25X^2 —> 25/25X^2 + 49/25X^2 = 74/25X^2

74/25X^2 = 1850/49 divide by 74 both sides

X^2/25 = 25/49 multiply by 25

X^2 = 625/49 square root both sides

X= + or - 25/7= + or - 3.57

Remember when you square root something, you create both a positive and negative solution.

Finally, we can replace big X with small x to get our 2 possibly x values for point D

X = x-7

3.57 = x - 7

and

-3.57= x - 7

x = 10.57 or x = 3.43

Lastly, we can find the y values that match these coordinates by plugging them into the original equation

y - 5 = -7/5(x - 7)

y - 5 = -7/5(10.57 - 7)

y - 5 = -7/5(3.57)

y - 5 = -5

y = 0

One possible coordinate point for D: (10.57,0)

y - 5 = -7/5(3.43 - 7)

y - 5 = -7/5(-3.57)

y - 5 = 5

y = 10

Second possible coordinate point for D: (3.43,10)