Question

1. Getting ∠AOC = 50º, Getting ∠ACO = 40º
1 1
2 2
+
2.
x 1 tan 30º
20 3 3
= = ⇒ x = 20 m
1
2
∴ AB = 21.7 m
1
2
3. 2(3k + 3) = (2k + 1) + (5k – 1)
1
2
⇒ k = 6
1
2
4. n(s) = 20, Multiples of 3 or 7 A: {3, 6, 9, 12, 15, 18, 7, 14} For n (A) = 8 ....
1
2
∴ Reqd. Probability =
8 2
or
20 5
1
2

Answer 1

Step-by-step explanation:

AOB is a straight line

∴∠AOC+∠COD+∠DOB=180

∘

⇒x+20

∘

+2x−20

∘

+60

∘

=180

∘

⇒3x=120

∘

⇒x=40

∘

∠COD=2x−20

∘

=(2×40

∘

)−20

∘

=60

∘

Answer 2

Step-by-step explanation:

∠AOC+BOD=2∠ABC+2∠BCD

=2∠ABC+∠BCD

=2∠BPD......

( since, angle formed on major arc is half of the angle formed on center)

∴∠AOC+∠BOD=2∠BPD

2∠BPD=50

o

+40

o

∠BPD=

2

90

o

=45

o