Answer:
Length of the rope = 15.6 ft.
Step-by-step explanation:
Given:
Height of the pole from the ground = 12 ft
Adjacent distance from the pole = 10 ft
We have to find the length of the rope attached by Scott to secure the pole.
According to the question :
In forma right angled triangle as shown in the figure.
We have to find the measure of the hypotenuse.
Length of rope = Hypotenuse measure
Let the length of the rope = "h" ft
Using Pythagoras formula.
⇒
⇒ Plugging the values.
⇒
⇒
⇒
⇒ ft
The length of the rope, rounded to the nearest tenth is 15.6 ft.
Answer:
Point R is at (−20, 10), a distance of 30 units from point Q
Step-by-step explanation:
Q has coordinates (-20,-20).
P has coordinates (10,-20)
Since point R is vertically above point Q, it will have the same x-coordinate as Q.
Let R have coordinates (-20,y).
It was given that;
.
The coordinates of R are (-20,10).
The dstance from Q is 30 units.
Answer:
Step-by-step explanation:
9) PQR Is an isosceles triangle
=> ∠PRQ = (180° - x)/2
PRS is an isosceles right triangle
=> ∠PRS = 45°
Have: ∠PRS + ∠PRQ = 115°
=>
=> 180 - x = (115 - 45).2 = 140
<=> x = 180 - 140 = 40
10) ABD is an isosceles right triangle => ∠ABD = 45°
BCD is an equilateral triangle => ∠CBD = 60°
have: x = ∠ABD + ∠CBD = 45° + 60° = 105°
11) have: x = y (2)
PQT is an isosceles triangle => ∠PQT = 180 - 70.2 = 40
QTS is an isosceles triangle => ∠TQS = 180 -2x
QRS is an isosceles triangle => ∠RSQ = y
have: 40 + 180 - 2x + y = 180 => 2x - y = 40 (1)
(1)(2) =>
=> x + y = 80
12) EFJ Is an equilateral triangle => ∠FJE = 60
∠FJE is the outer angle of the triangle FHJ but FHJ is an isosceles triangle
=> 60 = 2.∠JHF => ∠JHF = 30°
∠JHF is the outer angle of the triangle FHG
=> 30° = 2x
<=> x = 15°