help with graph

Question

 

 

A (0,0)

B (4,3)

C (7,0)

D (4,-4)

 

 

m∠A = 82 degrees (to the nearest degrees).  Find m∠B, m∠C, and m∠D to the nearest degree.

 

m∠B = ____

m∠C = ____

m∠D = ____

 

Thanks.

Answer 1

First, let's understand why ∠A = 82. The slope of line AB is (0-3)/(0-4) = -3/-4 = 3/4. That means to get from A to be, we go up 3 and right 4. The "up 3" line, the "right 4" line, and AB form a right triangle with legs 3 and 4. To find the measure of ∠CAB, then, we can use tangent:

 

tan(∠CAB) = opposite/adjacent

tan(∠CAB) = 3/4

∠CAB = tan^-1(3/4)

∠CAB ≈ 37

 

Similarly, AD has a slope of (0-(-4))/(0-4) = 4/-4 = -1. Again, we can use tangent to find ∠CAD:

 

tan(∠CAD) = opposite/adjacent

tan(∠CAD) = -1

∠CAD = tan^-1(-1)

∠CAD = 45

 

Together, that's a total of 45 + 37 = 82 = ∠A. Notice that, each time, we just took the inverse tangent of the line's slope.

 

 

 

Let's apply the same process to find ∠C.

 

First, the slope of BC is (3-0)/(4-7) = 3/-3 = -1. That means ∠BCA = tan^-1(-1) = 45.

 

The slope of CD is (0-(-4))/(7-4) = 4/3. That means ∠ACD = tan^-1(4/3) ≈ 53.

 

Together, that's a total of 45 + 53 = 98 = ∠ C.

 

 

 

We could do the same for ∠B and ∠D, but we already have everything we need to find those. Note that each angle is part of a triangle: ∠B is part of ΔABC and ∠D is part of ΔACD. Remember that the angles of a triangle add up to be 180.

 

We know tat ∠CAB = 37 and ∠BCA = 45, so ∠ B = 180 - 37 - 45 = 98.

 

Similarly, ∠CAD = 45 and ∠ACD = 53, so ∠ D = 180 - 45 - 53 = 82.

 

 

 

Let's check our work by adding our answers together, since every quadrilateral's angles should add up to 360:

 

∠A + ∠B + ∠C + ∠D = 360

82 + 98 + 98 + 82 = 360

 

That checks out!