Q4) Solve any two of the following subquestions
(8)
1) Prove the theorem of cyclic quadrilateral
2) A bird was flying in a line parallel to the ground from north to south at a
height of 2000 metres. Tom, standing in the middle of the field, first
observed the bird in the north at the angle of 30°. After 3 minutes, he
again observed it in the south at an angle of 45°. Find the speed of the
bird in kilometres per hour (V3 = 1.73
Answers
1:A rectangle has a length of 8 cm, and a width of 3 cm. Find its perimeter and area?
The ratio between the diagonals and the sides can be defined and is known as the Cyclic quadrilateral theorem. If there's a quadrilateral that is inscribed in a circle, then the product of the diagonals is equal to the sum of the product of its two pairs of opposite sides.
2: A bird was flying in a line parallel to the ground from north to south at a height of 2000 metres. Tom, standing in the middle of the field, first
observed the bird in the north at the angle of 30°. After 3 minutes, he
again observed it in the south at an angle of 45°. Find the speed of the
bird in kilometres per hour (V3 = 1.73)?
Answer: 109.20
Step-by-step explanation: a bird is flying to the north at an angle of elevation of 30°.
Using tan 30°=opp/adj
=1/√3=2000/BC
BC=2000√3m-------------(1)
Now, the bird fly to south at an angle of elevation of 45°
= tan45°=opp/adj
=1=2000/CD
=CD=2000m----------------(2)
Therefore,distance=BC+CD
2000√3+2000 =2000(√3+1)m
But we want the answer in km
2000(√3+1)/1000 = 2(√3+1)km
Bird takes 3min=3/60hr
Therefore,speed=distance/time
=2(√3+1)/3/60
=2(√3+1)×60/3
=20(2√3+2)
=40√3+40. (√3=1.73)
=40×1.73+40
=69.20+40
=109.20km/hr.
Answer:
1:A rectangle has a length of 8 cm, and a width of 3 cm. Find its perimeter and area?
The ratio between the diagonals and the sides can be defined and is known as the Cyclic quadrilateral theorem. If there's a quadrilateral that is inscribed in a circle, then the product of the diagonals is equal to the sum of the product of its two pairs of opposite sides.
2: A bird was flying in a line parallel to the ground from north to south at a height of 2000 metres. Tom, standing in the middle of the field, first
observed the bird in the north at the angle of 30°. After 3 minutes, he
again observed it in the south at an angle of 45°. Find the speed of the
bird in kilometres per hour (V3 = 1.73)?
Answer: 109.20
Step-by-step explanation: a bird is flying to the north at an angle of elevation of 30°.
Using tan 30°=opp/adj
=1/√3=2000/BC
BC=2000√3m-------------(1)
Now, the bird fly to south at an angle of elevation of 45°
= tan45°=opp/adj
=1=2000/CD
=CD=2000m----------------(2)
Therefore,distance=BC+CD
2000√3+2000 =2000(√3+1)m
But we want the answer in km
2000(√3+1)/1000 = 2(√3+1)km
Bird takes 3min=3/60hr
Therefore,speed=distance/time
=2(√3+1)/3/60
=2(√3+1)×60/3
=20(2√3+2)
=40√3+40. (√3=1.73)
=40×1.73+40
=69.20+40
=109.20km/hr.