Hi!
When you use postulates and theorems, you need to make sure to only use the given information that you know. Look for the given statements, and congruence marks on the figure. Those are also considered given.
By looking, you are given an angle and a side. The side comes first. SU=TV.
So, that makes it so Side-Angle-Side would be the best option.
I hope this helps!
Answer:
389 feets apart
Step-by-step explanation:
Given that:
Building A
Height = 400feets
Angle of elevation = 70°
Building B:
Height = 300 feets
Angle of elevation = 52°
Distance from the foot of building A to where Jonny is standing = x (see attached picture)
Using trigonometry :
Tanθ = opposite / Adjacent
Tan 70 = 400/ x
2.7474774x = 400
x = 145.58809 feets
Distance from the foot of building B to where Jonny is standing = y (see attached picture)
Using trigonometry :
Tanθ = opposite / Adjacent
Tan 52 = 300/ y
1.2799416y = 300
y = 145.59
y = 234.38568 feets
x + y = distance between the two buildings
(145.58809 + 234.38568) feets
= 379.97377
= 380 feets apart
The Associative Property allows you to "regroup" addition and multiplication problems. You can group this problem in two other ways,
(8 + 4) + 3 and (8 + 3) + 4.
Answer:
In the given figure the point on segment PQ is twice as from P as from Q is. What is the point? Ans is (2,1).
Step-by-step explanation:
There is really no need to use any quadratics or roots.
( Consider the same problem on the plain number line first. )
How do you find the number between 2 and 5 which is twice as far from 2 as from 5?
You take their difference, which is 3. Now splitting this distance by ratio 2:1 means the first distance is two thirds, the second is one third, so we get
4=2+23(5−2)
It works completely the same with geometric points (using vector operations), just linear interpolation: Call the result R, then
R=P+23(Q−P)
so in your case we get
R=(0,−1)+23(3,3)=(2,1)
Why does this work for 2D-distances as well, even if there seem to be roots involved? Because vector length behaves linearly after all! (meaning |t⋅a⃗ |=t|a⃗ | for any positive scalar t)
Edit: We'll try to divide a distance s into parts a and b such that a is twice as long as b. So it's a=2b and we get
s=a+b=2b+b=3b
⇔b=13s⇒a=23s
