Answer:
∠RPQ = 27
Step-by-step explanation:
In ΔSRQ,
∠R = 90
∠SQR = 36°
∠R + ∠SQR + ∠RSQ = 180 {Angle sum property of triangle}
90 + 36 + ∠RSQ = 180
126 + ∠RSQ = 180
∠RSQ = 180 - 126
∠RSQ = 54°
∠PSQ +∠RSQ = 180 {Linear pair}
∠PSQ + 54 = 180
∠PSQ = 180 - 54
∠PSQ = 126
In ΔPSQ,
SQ = PS ,
So, ∠SQP = ∠SPQ {Angles opposite to equal sides are equal}
∠SQP = ∠SPQ =x
∠PSQ + x +x = 180 {Angle sum property of triangle}
126 + 2x = 180
2x = 180 - 126
2x = 54
x = 54/2
x = 27
∠RPQ = 27°
distance is speed multiply by time, from speed equals distance divided by time so we have 100 multiply by 6 which is 600km
The artistic crop isn't helpful; it cuts off some vertex names.
The circumcenter H is the meet of the perpendicular bisectors of the sides, helpfully drawn. We have right triangle ELH, right angle L, so
EH² = HL² + EL²
EL = √(EH² - HL²) = √(5.06²-2.74²) ≈ 4.2539393507665339
Since HL is a perpendicular bisector of EF, we get congruent segments FL=EL.
Answer: 4.25
P = 2(L + W)
P = 34
L = W + 5
34 = 2(W + 5 + W)
34 = 2(2W + 5)
34 = 4W + 10
34 - 10 = 4W
24 = 4W
24/4 = W
6 = W <==== the width is 6 ft
L = W + 5
L = 6 + 5
L = 11 <==== the length is 11 ft
Answer:
the answer is C
Step-by-step explanation: