CBSE Class 10 Math Standard Sample Paper 2022 – 23 Solutions
CBSE Class 10 Math Standard Sample Paper 2022: Central Board of Secondary Education(CBSE) has released the CBSE Class 10 Math Standard Sample Paper 2022 – 23 on its official website on 16th September 2022.
1.) Let a and b be two positive integers such that a = p3q4 and b = p2q3 , where p and q are
prime numbers. If HCF(a,b) = pmqn and LCM(a,b) = prqs, then (m+n)(r+s)=
(a) 15 (b) 30 (c) 35 (d) 72
Ans- (c) 35
2.) Let p be a prime number. The quadratic equation having its roots as factors of p is
(a) x2 –px +p=0 (b) x2–(p+1)x +p=0 (c) x2+(p+1)x +p=0 (d) x2 –px+p+1=0
Ans- (b) x2–(p+1)x +p=0
3.) If α and β are the zeros of a polynomial f(x) = px2 – 2x + 3p and α + β = αβ, then p is
(a)-2/3 (b) 2/3 (c) 1/3 (d) -1/3
Ans- (b) 2/3
4.) If the system of equations 3x+y =1 and (2k-1)x +(k-1)y =2k+1 is inconsistent, then k =
(a) -1 (b) 0 (c) 1 (d) 2
Ans – (d) 2
5.) If the vertices of a parallelogram PQRS taken in order are P(3,4), Q(-2,3) and R(-3,-2), then the coordinates of its fourth vertex S are
(a) (-2,-1) (b) (-2,-3) (c) (2,-1) (d) (1,2)
Ans- (c) (2,-1)
6.) ∆ABC~∆PQR. If AM and PN are altitudes of ∆ABC and ∆PQR respectively and AB2: PQ2 = 4 : 9, then AM: PN =
(a) 3:2 (b) 16:81 (c) 4:9 (d) 2:3
Ans – (d) 2:3
7.) If x tan 60°cos60°= sin60°cot60°, then x =
(a) cos30° (b) tan30° (c) sin30° (d) cot30°
Ans – (b) tan30°
8.) If sinθ + cosθ = √2, then tanθ + cot θ =
(a) 1 (b) 2 (c) 3 (d) 4
Ans- (b) 2
9.) In the given figure, DE ∥ BC, AE = a units, EC =b units, DE =x units and BC = y units. Which of the following is true?
(a) x= a+b/ (b) y= ax/a+b (c) x= ay/a+b d) x/y =a/b
Ans – (c) x= ay/a+b
10.) ABCD is a trapezium with AD ∥ BC and AD = 4cm. If the diagonals AC and BD intersect each other at O such that AO/OC = DO/OB =1/2, then BC =
(a) 6cm (b) 7cm (c) 8cm (d) 9cm
Ans – (c) 8cm
11.) If two tangents inclined at an angle of 60ᵒ are drawn to a circle of radius 3cm, then the length of each tangent is equal to
a) 3√3 cm (b) 3cm (c) 6cm (d) 3√3cm
Ans – (d) 3√3cm
12.) The area of the circle that can be inscribed in a square of 6cm is
(a) 36π cm2 (b) 18π cm2 (c) 12 π cm2 (d) 9π cm2
Ans – (d) 9π cm2
13.) The sum of the length, breadth and height of a cuboid is 6√3cm and the length of its diagonal is 2√3cm. The total surface area of the cuboid is
(a) 48 cm2 (b) 72 cm2 (c) 96 cm2 (d) 108 cm2
Ans – (c) 96 cm2
14.) If the difference of Mode and Median of a data is 24, then the difference of median and mean is
(a) 8 (b) 12 (c) 24 (d) 36
Ans – (b) 12
15.) The number of revolutions made by a circular wheel of radius 0.25m in rolling a distance of 11km is
(a) 2800 (b) 4000 (c) 5500 (d) 7000
Ans – (d) 7000
16.) For the following distribution,
Class | 0-5 | 5-10 | 10-15 | 15-20 | 20-25 |
Frequency | 10 | 15 | 12 | 20 | 9 |
the sum of the lower limits of the median and modal class is
(a) 15 (b) 25 (c) 30 (d) 35
Ans – (b) 25
17.) Two dice are rolled simultaneously. What is the probability that 6 will come up at least once?
(a)1/6 (b) 7/36 (c) 11/36 (d) 13/36
Ans – (c) 11/36
18.) If 5 tanβ =4, then 5 sinβ – 2 cosβ /5 sinβ + 2 cosβ =
(a) 1/3 (b) 2/5 (c) 3/5 (d) 6
Ans – (a) 1/3
19.) DIRECTION: In the question number 19 and 20, a statement of assertion (A) is followed by a statement of Reason (R).
Choose the correct option
Statement A (Assertion): If product of two numbers is 5780 and their HCF is 17, then their LCM is 340
Statement R( Reason) : HCF is always a factor of LCM
a.) Both assertion (A) and reason (R) are true and reason (R) is the correct explanation of assertion (A)
b.) Both assertion (A) and reason (R) are true and reason (R) is not the correct explanation of assertion (A)
c.) Assertion (A) is true but reason (R) is false.
d.) Assertion (A) is false but reason (R) is true.
Ans – b.) Both assertion (A) and reason (R) are true and reason (R) is not the correct explanation of assertion (A)
20.) Statement A (Assertion): If the co-ordinates of the mid-points of the sides AB and AC of ∆ABC are D(3,5) and E(-3,-3) respectively, then BC = 20 units
Statement R( Reason) : The line joining the mid points of two sides of a triangle is parallel to the third side and equal to half of it.
a.) Both assertion (A) and reason (R) are true and reason (R) is the correct explanation of assertion (A)
b.) Both assertion (A) and reason (R) are true and reason (R) is not the correct explanation of assertion (A)
c.) Assertion (A) is true but reason(R) is false.
d.) Assertion (A) is false but reason(R) is true.
Ans – (a) Both assertion (A) and reason (R) are true and reason (R) is the correct explanation of assertion (A)
21.) If 49x+51y= 499, 51 x+49 y= 501, then find the value of x and y
Ans – Adding the two equations and dividing by 10, we get : x+y = 10 Subtracting the two equations and dividing by -2, we get : x-y =1
Solving these two new equations, we get, x = 11/2
22.) In the given figure below, AD/AE = AC/BD Show that ∆ BAE~ ∆CAD .
Ans – In ΔABC, ∠1 = ∠2 ∴ AB = BD ………………(i) Given, AD/AE = AC/BD Using equation (i), we get AD/AE = AC/AB ……………….(ii) In ΔBAE and ΔCAD, by equation (ii), AC/AB = AD/AE ∠A= ∠A (common) ∴ ΔBAE ~ ΔCAD [By SAS similarity criterion]
23.) In the given figure, O is the centre of circle. Find ∠AQB, given that PA and PB are tangents to the circle and ∠APB= 75°.
Ans – ∠PAO = ∠ PBO = 90° ( angle b/w radius and tangent) ∠AOB = 105° (By angle sum property of a triangle) ∠AQB = ½ x105° = 52.5° (Angle at the remaining part of the circle is half the angle subtended by the arc at the centre)
24.) The length of the minute hand of a clock is 6cm. Find the area swept by it when it moves from 7:05 p.m. to 7:40 p.m.
OR
In the given figure, arcs have been drawn of radius 7cm each with vertices A, B, C and D of quadrilateral ABCD as centres. Find the area of the shaded region.
We know that, in 60 minutes, the tip of minute hand moves 360° In 1 minute, it will move =360°/60 = 6°
∴ From 7 : 05 pm to 7: 40 pm i.e. 35 min, it will move through = 35 × 6° = 210°
∴ Area ofswept by the minute hand in 35 min = Area of sector with sectorial angle θ of 210° and radius of 6 cm =
= 210/360 x π x 6^2
= 7/12 x 22/7 x 6 x 6
= 66cm2
Or
Let the measure of ∠A, ∠B, ∠C and ∠D be θ₁, θ₂, θ₃ and θ₄ respectively Required area = Area ofsector with centre A + Area ofsector with centre B + Area ofsector with centre C + Area ofsector with centre D
= ₁/360 x π x 7 2 + ₂/360 x π x 72+ ₃/360 x π x 72 + ₄/360 x π x 72
= (₁ + ₂ + ₃ + ₄)/360 x π x 72
= () 360 x /7 x 7x 7 ( By angle sum property of a triangle)
= 154 cm2
25.) If sin(A+B) =1 and cos(A-B)= √3/2, 0°< A+B ≤ 90° and A> B, then find the measures of angles A and B.
Ans – sin(A+B) =1 = sin 90, so A+B = 90……………….(i)
cos(A-B)= √3/2 = cos 30, so A-B= 30……………(ii)
From (i) & (ii) ∠A = 60° And ∠B = 30°
OR
Find an acute angle θ when
cosθ − sin θ/cosθ+sin θ = 1−√3/1+√3
Dividing the numerator and denominator of LHS by cosθ, we get
1 − tan θ/1+tan θ = 1−√3/1+√3
Which on simplification (or comparison) gives tanθ = √3
Or θ= 60°
SECTION C
Section C consists of 6 questions of 3 marks each.
26.) Given that √3 is irrational, prove that 5 + 2√3 is irrational.
Let us assume 5 + 2√3 is rational, then it must be in the form of p/q where p and q are co-prime integers and q ≠ 0
i.e 5 + 2√3 = p/q
So √3 = −5 2 ……………………(i)
Since p, q, 5 and 2 are integers and q ≠ 0, HS of equation (i) is rational. But LHS of (i) is √3 which is irrational. This is not possible.
This contradiction has arisen due to our wrong assumption that 5 + 2√3 is rational. So, 5 + 2√3 is irrational.
27.) If the zeroes of the polynomial x2 +px +q are double in value to the zeroes of the polynomial 2x2 -5x -3, then find the values of p and q.
Ans – Let α and β be the zeros of the polynomial 2×2 -5x -3
Then α + β = 5/2
And αβ = -3/2.
Let 2α and 2β be the zeros x2 + px +q Then 2α + 2β = -p 2(α + β) = -p 2 x 5/2 =-p
So p = -5 And 2α x 2β = q 4 αβ = q
So q = 4 x-3/2 = -6
28.) A train covered a certain distance at a uniform speed. If the train would have been 6 km/h faster, it would have taken 4 hours less than the scheduled time. And, if the train were slower by 6 km/hr ; it would have taken 6 hours more than the scheduled time. Find the length of the journey.
Ans – Let the actual speed of the train be x km/hr and let the actual time taken be y hours. Distance covered is xy km
If the speed is increased by 6 km/hr, then time of journey isreduced by 4 hoursi.e.,
when speed is(x+6)km/hr, time of journey is(y−4) hours.
∴ Distance covered =(x+6)(y−4) ⇒xy=(x+6)(y−4) ⇒−4x+6y−24=0 ⇒−2x+3y−12=0 …….(i)
Similarly xy=(x−6)(y+6) ⇒6x−6y−36=0 ⇒x−y−6=0 …………(ii)
Solving (i) and (ii) we get x=30 and y=24
Putting the values of x and y in equation (i),
we obtain Distance =(30×24)km =720km. Hence, the length of the journey is 720km.
OR
Anuj had some chocolates, and he divided them into two lots A and B. He sold the first lot at the rate of ₹2 for 3 chocolates and the second lot at the rate of ₹1 per chocolate, and got a total of ₹400. If he had sold the first lot at the rate of ₹1 per chocolate, and the second lot at the rate of ₹4 for 5 chocolates, his total collection would have been ₹460.
Find the total number of chocolates he had.
Ans – Let the number of chocolates in lot A be x
And let the number of chocolates in lot B be y
∴ total number of chocolates =x+y
Price of 1 chocolate = ₹ 2/3 , so for x chocolates = x and price of y chocolates at the rate of ₹ 1 per chocolate =y.
∴ by the given condition x +y=400 ⇒2x+3y=1200 …………..(i)
Similarly x+ y = 460 ⇒5x+4y=2300 …….. (ii)
Solving (i) and (ii) we get x=300 and y=200 ∴x+y=300+200=500
So, Anuj had 500 chocolates.
29.) Prove the following that-
tan3θ + cot3θ = secθcosecθ – 2 sinθcosθ
1+ tan2θ 1+ cot2θ
LHS : sin3 θ/ cos3 θ + cos3 θ/ sin3 θ 1+ sin2 θ/cos2 θ 1+ cos2 θ/ sin2 θ
= sin3 θ/ cos3 θ + cos 3 θ/ sin3 θ (cos2 θ + sin2 θ)/cos2 θ (sin2 θ + cos2 θ)/ sin2 θ
= sin3 θ + cos3 θ cosθ sinθ = sin4 θ + cos4 θ cosθsinθ
= (sin2 θ + cos2 θ)2 – 2 sin2 θcos2 θ cosθsinθ
= 1 – 2 sin2 θcos2 θ cosθsinθ = 1 – 2 sin2 θcos2 θ cosθsinθ cosθsinθ
= secθcosecθ – 2sinθcosθ
= RHS
30.) Prove that a parallelogram circumscribing a circle is a rhombus
In the figure XY and X’Y’ are two parallel tangents to a circle with centre O and another tangent AB with point of contact C interesting XY at A and X’Y’ at B, what is the measure of AOB.
Let ABCD be the rhombus circumscribing the circle with centre O, such that AB, BC, CD and DA touch the circle at points P, Q, R and S respectively. We know that the tangents drawn to a circle from an exterior point are equal in length.
∴ AP = AS………….(1)
BP = BQ……………(2)
CR = CQ ……………(3)
DR = DS……………(4).
Adding (1), (2), (3) and (4) we get
AP+BP+CR+DR = AS+BQ+CQ+DS (AP+BP) + (CR+DR) = (AS+DS) + (BQ+CQ)
∴ AB+CD=AD+BC———–(5)
Since AB=DC and AD=BC (opposite sides of parallelogram ABCD)
putting in (5) we get, 2AB=2AD or AB = AD. ∴ AB=BC=DC=AD
Since a parallelogram with equal adjacent sides is a rhombus,
so ABCD is a rhombus
OR
Join OC
In Δ OPA and Δ OCA
OP = OC (radii of same circle)
PA = CA (length of two tangents from an external point)
AO = AO (Common)
Therefore, Δ OPA ≅ Δ OCA (By SSS congruency criterion)
Hence, ∠ 1 = ∠ 2 (CPCT)
Similarly ∠ 3 = ∠ 4 ∠PAB + ∠QBA =180°(co interior angles are supplementary as XY∥X’Y’) 2∠2 + 2∠4 = 180° ∠2 + ∠4 = 90°————————-(1)
∠2 + ∠4 +∠AOB = 180° (Angle sum property) Using (1), we get, ∠AOB = 90°
31.) Two coins are tossed simultaneously. What is the probability of getting
(i) At least one head?
- At most one tail?
- A head and a tail?
Ans- (i) P (At least one head) = 3/4 (ii) P(At most one tail) = 3/4 (iii) P(A head and a tail) = 2/4 = 1 2
SECTION D
Section D consists of 4 questions of 5 marks each.
32.) To fill a swimming pool two pipes are used. If the pipe of larger diameter used for 4 hours and the pipe of smaller diameter for 9 hours, only half of the pool can be filled. Find, how long it would take for each pipe to fill the pool separately, if the pipe of smaller diameter takes 10 hours more than the pipe of larger diameter to fill the pool?
Ans – Let the time taken by larger pipe alone to fill the tank= x hours Therefore, the time taken by the smaller pipe = x+10 hours Water filled by larger pipe running for 4 hours = 4/ litres Water filled by smaller pipe running for 9 hours = 9/+10 litres
We know that 4/ + 9/+10 = 1/2
Which on simplification gives: x2−16x−80=0
x2−20x + 4x−80=0
x(x-20) + 4(x-20)= 0
(x +4)(x-20)= 0
x=- 4, 20 x cannot be negative.
Thus, x=20 x+10= 30 Larger pipe would alone fill the tank in 20 hours and smaller pipe would fill the tank alone in 30 hours.
OR
In a flight of 600km, an aircraft was slowed down due to bad weather. Its average speed for the trip was reduced by 200 km/hr from its usual speed and the time of the flight increased by 30 min. Find the scheduled duration of the flight.
Ans – Let the usual speed of plane be x km/hr and the reduced speed of the plane be (x-200) km/hr
Distance =600 km [Given]
According to the question, (time taken at reduced speed) – (Schedule time) = 30 minutes = 0.5 hours.
600/−200 – 600/ = 1/2
Which on simplification gives:
x2 – 200x−240000=0
x2 -600x + 400x −240000=0
x(x- 600) + 400( x-600) = 0
(x-600)(x+400) =0 x=600
or x=−400
But speed cannot be negative. ∴ The usual speed is 600 km/hr and the scheduled duration of the flight is 600 600 =1hour
33.) Prove that if a line is drawn parallel to one side of a triangle intersecting the other two sides in distinct points, then the other two sides are divided in the same ratio. Using the above theorem prove that a line through the point of intersection of the diagonals and parallel to the base of the trapezium divides the non parallel sides in the same ratio.
For the Theorem : Given, To prove, Construction and figure Proof
Let ABCD be a trapezium DC∥AB and EF is a line parallel to AB and hence to DC.
To prove : / = /
Construction :Join AC, meeting EF in G.
Proof :
In △ABC, we have
GF∥AB
CG/GA=CF/FB [By BPT] ……(1)
In △ADC, we have EG∥DC ( EF ∥AB & AB ∥DC)
DE/EA= CG/GA [By BPT] …..(2)
From (1) & (2), we get, / = /B
34.) Due to heavy floods in a state, thousands were rendered homeless. 50 schools collectively decided to provide place and the canvas for 1500 tents and share the whole expenditure equally. The lower part of each tent is cylindrical with base radius 2.8 m and height 3.5 m and the upper part is conical with the same base radius, but of height 2.1 m. If the canvas used to make the tents costs ₹120 per m2 , find the amount shared by each school to set up the tents.
Ans – Radius of the base of cylinder (r) = 2.8 m = Radius of the base of the cone (r)
Height of the cylinder (h)=3.5 m Height of the cone (H)=2.1 m.
Slant height of conical part (l)=√r2+H2
= √(2.8)2+(2.1)2
= √7.84+4.41 = √12.25
= 3.5 m
Area of canvas used to make tent = CSA of cylinder + CSA of cone
= 2×π×2.8×3.5 + π×2.8×3.5
= 61.6+30.8
= 92.4m2
Cost of 1500 tents at ₹120 per sq.m = 1500×120×92.4 = 16,632,000 Share of each school to set up the tents = 16632000/50 = ₹332,640
OR
There are two identical solid cubical boxes of side 7cm. From the top face of the first cube a hemisphere of diameter equal to the side of the cube is scooped out. This hemisphere is inverted and placed on the top of the second cube’s surface to form a dome. Find (i) the ratio of the total surface area of the two new solids formed (ii) volume of each new solid formed.
Ans – (i) SA for first new solid (S₁):
6×7×7 + 2 π ×3.52- π ×3.52
= 294 + 77 – 38.5
= 332.5cm2
SA for second new solid (S₂):
6×7×7 + 2 π ×3.52
– π ×3.52
= 294 + 77 – 38.5
= 332.5 cm2
So S₁: S₂ = 1:1
(ii) Volume for first new solid (V₁)= 7×7×7 – 2/3 π ×3.53
= 343 – 539/6 = 1519/6 cm3
Volume for second new solid (V₂)= 7×7×7 + 2 3 π ×3.53
= 343 + 539/6 = 2597/6 cm3
35.) The median of the following data is 525. Find the values of x and y, if the total frequency is 100
Class interval | Frequency |
0-100 | 2 |
100-200 | 5 |
200-300 | x |
300-400 | 12 |
400-500 | 17 |
500-600 | 20 |
600-700 | y |
700-800 | 9 |
800-900 | 7 |
900-1000 | 4 |
Median = 525, so Median Class = 500 – 600
Class interval | Frequency | Cumulative Frequency |
0-100 | 2 | 2 |
100-200 | 5 | 7 |
200-300 | x | 7+x |
300-400 | 12 | 19+x |
400-500 | 17 | 36+x |
500-600 | 20 | 56+x |
600-700 | y | 56+x+y |
700-800 | 9 | 65+x+y |
800-900 | 7 | 72+x+y |
900-1000 | 4 | 76+x+y |
76+x+y=100⇒x+y=24 ….(i)
Median = l + n/2 –cf/f x h
Since, l=500, h=100, f=20, cf=36+x and n=100
Therefore, putting the value in the Median formula, we get;
525 = 500 + 50−(36+x)/20 x 100
so x = 9 y = 24 – x (from eq.i) y = 24 – 9 = 15
Therefore, the value of x = 9
and y = 15.
SECTION E
Case study based questions are compulsory.
36.) i.) A tiling or tessellation of a flat surface is the covering of a plane using one or more geometric shapes, called tiles, with no overlaps and no gaps. Historically, tessellations were used in ancient Rome and in Islamic art. You may find tessellation patterns on floors, walls, paintings etc. Shown below is a tiled floor in the archaeological Museum of Seville, made using squares, triangles and hexagons.
Ans – (i) B(1,2), F(-2,9)
BF² = ( -2-1)²+ ( 9-2)²
= ( -3)²+ ( 7)² = 9 + 49
= 58 So, BF = √58 units
ii) A craftsman thought of making a floor pattern after being inspired by the above design. To ensure accuracy in his work, he made the pattern on the Cartesian plane. He used regular octagons, squares and triangles for his floor tessellation pattern
Use the above figure to answer the questions that follow:
i.) What is the length of the line segment joining points B and F?
ii.) The centre ‘Z’ of the figure will be the point of intersection of the diagonals of quadrilateral WXOP. Then what are the coordinates of Z?
iii.) What are the coordinates of the point on y axis equidistant from A and G?
OR
What is the area of Trapezium AFGH?
Ans – A(-2,2), G(-4,7)
Let the point on y-axis be Z(0,y) AZ² = GZ²
( 0+2)² + ( y-2)² = ( 0+4)² + ( y-7)²
( 2)² + y² + 4 -4y= (4)²+ y² + 49 -14y
8-4y = 65-14y 10y= 57
So, y= 5.7 i.e. the required point is (0, 5.7)
Or,
A(-2,2), F(-2,9), G(-4,7), H(-4,4)
Clearly GH = 7-4=3units
AF = 9-2=7 units
So, height of the trapezium AFGH = 2 units
So, area of AFGH = 1 2 (AF + GH) x height = 1 2 (7+3) x 2 = 10 sq. units
37.) The school auditorium was to be constructed to accommodate at least 1500 people. The chairs are to be placed in concentric circular arrangement in such a way that each succeeding circular row has 10 seats more than the previous one.
Ans – (i) Since each row is increasing by 10 seats,
so it is an AP with first term a= 30, and common difference d=10.
So number of seats in 10th row = 10 = a+ 9d
= 30 + 9×10 = 120
(ii) Sn = n 2 ( 2a + (n-1)d) 1500 = n 2 ( 2 × 30 + (n-1)10) 3000 = 50n + 10n2 n 2 +5n -300 =0 n 2 + 20n -15n – 300 =0 (n+20) (n-15) =0
Rejecting the negative value, n= 15
(i) If the first circular row has 30 seats, how many seats will be there in the 10th row?
(ii) For 1500 seats in the auditorium, how many rows need to be there?
iii.) If 1500 seats are to be arranged in the auditorium, how many seats are still left to be put after 10th row?
iv.) If there were 17 rows in the auditorium, how many seats will be there in the middle row?
Ans – No. of seats already put up to the 10th row = S10 S10 = 10 2 {2 × 30 + (10-1)10)}
= 5(60 + 90) = 750
So, the number of seats still required to be put are 1500 -750 = 750
(iii) If no. of rows =17 then the middle row is the 9 th row 8 = a+ 8d = 30 + 80 = 110 seats
38.) We all have seen the airplanes flying in the sky but might have not thought of how they actually reach the correct destination. Air Traffic Control (ATC) is a service provided by ground-based air traffic controllers who direct aircraft on the ground and through a given section of controlled airspace, and can provide advisory services to aircraft in non-controlled airspace. Actually, all this air traffic is managed and regulated by using various concepts based on coordinate geometry and trigonometry.
At a given instance, ATC finds that the angle of elevation of an airplane from a point on the ground is 60°. After a flight of 30 seconds, it is observed that the angle of elevation changes to 30°. The height of the plane remains constantly as 3000√3 m. Use the above information to answer the questions that follow-
- Draw a neat labelled figure to show the above situation diagrammatically.
- What is the distance travelled by the plane in 30 seconds?
OR
Keeping the height constant, during the above flight, it was observed that after 15(√3 -1) seconds, the angle of elevation changed
Ans – P and Q are the two positions of the plane flying at a height of 3000√3m. A is the point of observation.
(ii) In △ PAB, tan60° =PB/AB
Or √3 = 3000√3/ AB
So AB=3000m
tan30°= QC/AC
1/√3= 3000√3 /
AC AC = 9000m
distance covered = 9000- 3000 = 6000 m.
or,
In △ PAB, tan60° =PB/AB Or √3 = 3000√3/ AB So AB=3000m tan45° = RD/AD 1= 3000√3 / AD
AD = 3000√3 m
distance covered = 3000√3 – 3000
= 3000(√3 -1)m.
(iii) speed = 6000/ 30 = 200 m/s = 200 x 3600/1000 = 720km/hr
Alternatively: speed = 3000(√3 −1) 15(√3 −1) = 200 m/s = 200 x 3600/1000 = 720km/hr
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