Find, correct to the nearest degree, the three angles of the triangle with the vertices d(0,1,1), e( 2, 4,3) − , and f(1, 2, 1)
Ksju [112]
Well, here's one way to do it at least...
<span>For reference, let 'a' be the side opposite A (segment BC), 'b' be the side opposite B (segment AC) and 'c' be the side opposite C (segment AB). </span>
<span>Let P=(4,0) be the projection of B onto the x-axis. </span>
<span>Let Q=(-3,0) be the projection of C onto the x-axis. </span>
<span>Look at the angle QAC. It has tangent = 5/4 (do you see why?), so angle A is atan(5/4). </span>
<span>Likewise, angle PAB has tangent = 6/3 = 2, so angle PAB is atan(2). </span>
<span>Angle A, then, is 180 - atan(5/4) - atan(2) = 65.225. One down, two to go. </span>
<span>||b|| = sqrt(41) (use Pythagorian Theorum on triangle AQC) </span>
<span>||c|| = sqrt(45) (use Pythagorian Theorum on triangle APB) </span>
<span>Using the Law of Cosines... </span>
<span>||a||^2 = ||b||^2 + ||c||^2 - 2(||b||)(||c||)cos(A) </span>
<span>||a||^2 = 41 + 45 - 2(sqrt(41))(sqrt(45))(.4191) </span>
<span>||a||^2 = 86 - 36 </span>
<span>||a||^2 = 50 </span>
<span>||a|| = sqrt(50) </span>
<span>Now apply the Law of Sines to find the other two angles. </span>
<span>||b|| / sin(B) = ||a|| / sin(A) </span>
<span>sqrt(41) / sin(B) = sqrt(50) / .9080 </span>
<span>(.9080)sqrt(41) / sqrt(50) = sin(B) </span>
<span>.8222 = sin(B) </span>
<span>asin(.8222) = B </span>
<span>55.305 = B </span>
<span>Two down, one to go... </span>
<span>||c|| / sin(C) = ||a|| / sin(A) </span>
<span>sqrt(45) / sin(C) = sqrt(50) / .9080 </span>
<span>(.9080)sqrt(45) / sqrt(50) = sin(C) </span>
<span>.8614 = sin(C) </span>
<span>asin(.8614) = C </span>
<span>59.470 = C </span>
<span>So your three angles are: </span>
<span>A = 65.225 </span>
<span>B = 55.305 </span>
<span>C = 59.470 </span>
There is no option of the box plots, so I have created a version that would represent this data.
To make the box plot you will need the lower extreme, lower quartile, median, upper quartile, and upper extreme.
Please see the attached picture.
Given: AD ≅ BC and AD ∥ BC
Prove: ABCD is a parallelogram.
Statements Reasons
1. AD ≅ BC; AD ∥ BC 1. given
2. ∠CAD and ∠ACB are alternate interior ∠s 2. definition of alternate interior angles
3. ∠CAD ≅ ∠ACB 3. alternate interior angles are congruent
4. AC ≅ AC 4. reflexive property
5. △CAD ≅ △ACB 5. SAS congruency theorem
6. AB ≅ CD 6. Corresponding Parts of Congruent triangles are Congruent (CPCTC)
7. ABCD is a parallelogram 7. parallelogram side theorem
<span> </span>
Answer:
a)
b)d(t)=500t
c)
Step-by-step explanation:
d = Horizontal distance
s = the distance between the plane and the radar station
The horizontal distance (d), the one mile altitude, and s form a right triangle.
So, use Pythagoras theorem


a) 
(b) Express d as a function of the time t (in hours) that the plane has flown.

d(t)=500t
(c) Use composition to express s as a function of t.

using b

12 divided by 4 equals to 3
she paints 3 wall per hour
the answer is : the remaining wall will take her 1 hour to paint