Students can use CBSE Previous Year Question Papers Class 10 Maths with Solutions and CBSE Class 10 Maths Question Paper 2015 (Delhi) to familiarize themselves with the exam format and marking scheme.
CBSE Class 10 Maths Question Paper 2015 (Delhi) with Solutions
Time allowed: 3 hours
Maximum marks: 80
General Instructions:
Read the following instructions carefully and follow them:
- All questions are compulsory.
- This question paper contains of 30 questions.
- Question No. 1 – 6 in Section A are very short answer type questions carrying 1 mark each.
- Question No. 7 – 12 in Section B are short answer type questions carrying 2 marks each.
- Question No. 13 – 22 in Section C are long answer – I type questions carrying 3 marks each.
- Question No. 23 – 30 in Section D are long answer – II type questions carrying 4 marks each.
Section – A
Questions number 1 to 6 carry 1 mark each.
Question 1.
In ΔDEW, AB || EW. If AD = 4 cm, DE = 12 cm and DW = 24 cm, then find the value of DB.
Answer:
Let BD = x cm
then BW = (24 – x)cm, AE = 12 – 4 = 8 cm
In ΔDEW, AB || EW
\(\frac{\mathrm{AD}}{\mathrm{AE}}\) = \(\frac{\mathrm{BD}}{\mathrm{BW}}\)
AD _ BD AB||BW …[Thales’ Theorem
\(\frac{4}{8}\) = \(\frac{x}{24-x}\)
⇒ 8x = 96 – 4x
x = \(\frac{96}{12}\) = 8 cm ∴ DB = 8 cm
Question 2.
If √3 sin θ = cos θ, find the value of \(\frac{3 \cos ^2 \theta+2 \cos \theta}{3 \cos \theta+2}\).
Answer:
Given: √3 sin θ = cos θ
\(\frac{1}{\sqrt{3}}\) = \(\frac{\sin \theta}{\cos \theta}\) ⇒ tan θ = \(\frac{1}{\sqrt{3}}\)
tan θ = tan 30° ⇒ θ = 30° …….. (i)
Now, \(\frac{3 \cos ^2 \theta+2 \cos \theta}{3 \cos \theta+2}\) = \(\frac{\cos \theta(3 \cos \theta+2)}{(3 \cos \theta+2)}\)
= cos θ = cos 30° = \(\frac{\sqrt{3}}{2}\) …[From (i)
Question 3.
If x = – \(\frac{1}{2}\), is a solution of the quadratic equation 3x
2
+ 2kx – 3 = 0, find the value of k.
Answer:
The given quadratic equation can be written as, 3x
2
+ 2kx – 3 = 0
3(\(\frac{-1}{2}\))
2
+ 2k(\(\frac{-1}{2}\)) – 3 = 0 …[Putting x = \(\frac{-1}{2}\)
\(\frac{3}{4}\) – k – 3 = 0 ⇒ – k + \(\frac{3-12}{4}\) = 0
⇒ – k – \(\frac{9}{4}\) = 0 k = \(\frac{-9}{4}\)
Question 4.
The tops of two towers of height x and y, standing on level ground, subtend angles of 30° and 60° respectively at the centre of the line joining their feet, then find x : y.
Answer:
When base is same for both towers and their heights are given, i.e., x and y respectively. Let the base of towers be k.
tan 30° = \(\frac{x}{k}\)
x = k tan 30°
x = \(\frac{k}{\sqrt{3}}\) ……….. (i)
tan 60° = \(\frac{y}{k}\)
y = k tan 60°
y = k√3 ………… (ii)
From equations (i) and (ii),
\(\frac{x}{y}\) = \(\frac{\frac{k}{\sqrt{3}}}{k \sqrt{3}}\) = \(\frac{k}{\sqrt{3}}\) × \(\frac{1}{k \sqrt{3}}\) = \(\frac{1}{3}\) = 1 : 3
Question 5.
A letter of English alphabet is chosen at random. Determine the probability that the chosen letter is a consonant.
Answer:
Total English alphabets = 26
Number of consonants = 21
∴ P (letter is a consonant) = \(\frac{21}{26}\)
Question 6.
In Fig., PA and PB are tangents to the circle with centre O such that ∠APB = 50°. Write the measure of ∠OAB.
Answer:
PA = PB …[Tangents drawn from external point are equal
∠OAP = ∠OBP = 90°
∠OAB = ∠OBA ……(i) …[Angles opposite equal sides
∠OAP + ∠AOB + ∠OBP + ∠APB = 360° …[Quadratic rule
90° + ∠AOB + 90° + 50° = 360°
∠AOB = 360° – 230° = 130°
∠AOB + ∠OAB + ∠OBA = 180° …[Triangle rule
130° + 2∠OAB = 180° …[From (i) & (ii)
2∠OAB = 50° ∴ ∠OAB = 25°
Section – B
Questions number 7 to 12 carry 2 marks each.
Question 7.
Explain why (17 × 5 × 11 × 3 × 2 + 2 × 11) is a composite number?
Answer:
17 × 5 × 11 × 3 × 2 + 2 × 11 ………… (i)
= 2 × 11 × (17 × 5 × 3 + 1)
= 2 × 11 × (255 + 1) = 2 × 11 × 256
Number (i) is divisible by 2, 11 and 256 and it has more than 2 prime factors.
∴ (17 × 5 × 11 × 3 × 2 + 2 × 11) is a composite number.
Question 8.
X and Y are points on the sides AB and AC respectively of a triangle ABC such that \(\frac{\mathrm{AX}}{\mathrm{AB}}\) = \(\frac{1}{4}\), AY = 2 cm and YC = 6 cm. Find whether XY || BC or not.
Or
In Fig., AB is the diameter of a circle with centre O and AT is a tangent.
If ∠AOQ = 58°, find ∠ATQ.
Answer:
\(\frac{\mathrm{AX}}{\mathrm{AB}}\) = \(\frac{1}{4}\) ….[Given
AX = 1K, AB = 4K
∴ BX = AB – AX
= 4K – 1K = 3K
\(\frac{\mathrm{AX}}{\mathrm{AB}}\) = \(\frac{1 \mathrm{~K}}{3 \mathrm{~K}}\) = \(\frac{1}{3}\)
\(\frac{\mathrm{AY}}{\mathrm{YC}}\) = \(\frac{2}{6}\) = \(\frac{1}{3}\) …[each is 1/3
∴ XY||BC ..[By converse of Thales’ theorem
Or
∠ABQ = \(\frac{1}{2}\) ∠AOQ = \(\frac{58^{\circ}}{2}\) = 29°
∠BAT = 90° …[Tangent is ⊥ to the radius through the point of contact
∠ATQ = 180° – (∠ABQ + ∠BAT)
= 180° – (29° + 90°) = 180° – 119°
= 61°
Question 9.
Prove the identity : \(\frac{\sin ^3 \theta+\cos ^3 \theta}{\sin \theta+\cos \theta}\) = 1 – sin θ . cos θ.
Answer:
L.H.S = \(\frac{\sin ^3 \theta+\cos ^3 \theta}{\sin \theta+\cos \theta}\)
= \(\frac{(\sin \theta+\cos \theta)\left(\sin ^2 \theta+\cos ^2 \theta-\sin \theta \cos \theta\right)}{(\sin \theta+\cos \theta)}\) …[∵ a
3
+ b
3
= (a + b) (a
2
+ b
2
– ab)
= 1 – sin θ cos θ = R.H.S …[∵ sin
2
θ + cos
2
θ = 1
Question 10.
Solve the quadratic equation for x: 4x
2
– 4a
2
x + (a
4
– b
4
) = 0.
Answer:
The given quadratic equation can be written as,
4x
2
– 4a
2
x + (a
4
– b
4
) = 0
(4x
2
– 4a
2
x + a
4
) – b
4
= 0
(2x – a
2
)
2
– (b
2
)
2
= 0
∴ (2x – a
2
+ b
2
) (2x – a
2
– b
2
) = 0
⇒ (2x – a
2
+ b
2
) = 0 or (2x – a
2
– b
2
) = 0
⇒ x = \(\frac{a^2-b^2}{2}\) or x = \(\frac{a^2+b^2}{2}\)
Question 11.
Find the middle term of the A.P. 6,13, 20, …, 216.
Answer:
The given A.P. is 6,13, 20, …, 216
Let n be the number of terms,
d = 7, a = 6, a
n
= 216
a
n
= a + (n – 1 )d
∴ 216 = 6 + (n – 1).7
216 – 6 = (n – 1)7
\(\frac{210}{7}\) = n – 1 ⇒ 30 + 1 = n
⇒ n = 31
Middle term = (\(\frac{n+1}{2}\))
th
term
= (\(\frac{31+1}{2}\)) = (\(\frac{32}{2}\)) = 16
th
term of the A.P.
∴ a
16
= a + 15d = 6 + 15 × 7 = 111
Question 12.
If A(5, 2), B(2, -2) and C(-2, t) are the vertices of a right angled triangle with ∠B = 90°, then find the value of t.
Answer:
ABC is a right angled triangle,
∴ AC
2
= BC
2
+ AB
2
…(i) …[Pythagoras’ theorem
Using distance formula,
AB
2
= (5 – 2)
2
+ (2 + 2)
2
= 9 + 16 = 25
BC
2
– (2 + 2)
2
+ (t + 2)
2
= 16 + (t + 2)
2
AC
2
= (5 + 2)
2
+ (2 – t)
2
= 49 + (2 – t)
2
Putting values of AB
2
, AC
2
and BC
2
in equation (i), we get
49 + (2 – t)
2
= 16 + (t + 2)
2
+ 25
∴ 49 + (2 – t)
2
= 41 + (t + 2)
2
⇒ (t + 2)
2
– (2 – t)
2
= 8
⇒ (t
2
+ 4 + 4t – 4 – t
2
+ 4t)
⇒ 8t = 8 ⇒ t = 1
Section – C
Questions number 13 to 22 carry 3 marks each.
Question 13.
The length, breadth and height of a room are 8 m 50 cm, 6 m 25 cm and 4 m 75 cm respectively. Find the length of the longest rod that can measure the dimensions of the room exactly.
Answer:
To find the length of the longest rod,
We have to find HCF
L, Length = 8 m 50 cm = 850 cm = 2
1
× 5
2
× 17
B, Breadth = 6 m 25 cm = 625 cm = 5
4
H, Height = 4 m 75 cm = 475 cm = 5
2
× 19
∴ HCF of L, B and H is 5
2
= 25 cm
∴ Length of the longest rod = 25 cm
Question 14.
The sum of the digits of a two digit number is 8 and the difference between the number and that formed by reversing the digits is 18. Find the number.
Answer:
Let unit and tens digit be x and y.
Original number = 1x + 10y …….. (i)
Reversed number = 10x + 1y
According to question,
x + y = 8
⇒ y = 8 – x …… (ii)
1x + 10y – (10x + y) = 18
⇒ x + 10y – 10x – y = 18
⇒ 9y – 9x = 18
⇒ y – x = 2 …[Dividing both sides by 9
⇒ 8 – x – x = 2 …[From (ii)
⇒ 8 – 2 = 2x
⇒ 2x = 6 ⇒ x = 3
From (ii), y = 8 – 3 = 5
From (i), Original number = 3 + 10(5) = 53
Question 15.
In the Fig., EB ⊥ AC, BG ⊥ AE and CF ⊥ AE Prove that:
(a) ∆ABG ~ ∆DCB
(b) \(\frac{\mathrm{BC}}{\mathrm{BD}}\) = \(\frac{\mathrm{BE}}{\mathrm{BA}}\)
Answer:
Given: EB ⊥ AC, BG ⊥ AE and CF ⊥ AE.
To prove: (i) ΔABG ~ ΔDCB
(ii) \(\frac{\mathrm{BC}}{\mathrm{BD}}\) = \(\frac{\mathrm{BE}}{\mathrm{BA}}\)
Proof: (i) In ΔABG and ΔDCB
∠2 = ∠5 …[each 90°
∠6 = ∠4 …[corresponding angles
∴ ΔABG ~ ΔDCB (Hence Proved) …[By AA similarity
∠1 = ∠3 …[CPCT …[In – Δs, corresponding angles are equal
(ii) In ΔABE and ΔDBC
∠1 = ∠3 …[proved above
∠ABE = ∠5 …[each is 90°, EB ⊥ AC (Given)
ΔABE ~ ΔDBC …[By AA similarity
\(\frac{\mathrm{BC}}{\mathrm{BE}}\) = \(\frac{\mathrm{BD}}{\mathrm{BA}}\) …[In – Δs, corresponding sides are proportional
∴ \(\frac{\mathrm{BC}}{\mathrm{BD}}\) = \(\frac{\mathrm{BE}}{\mathrm{BA}}\) (Hence Proved)
Question 16.
If sin θ = \(\frac{12}{13}\) 0° < θ < 90°, find the value of: \(\frac{\sin ^2 \theta-\cos ^2 \theta}{2 \sin \theta \cdot \cos \theta} \times \frac{1}{\tan ^2 \theta}\)
Answer:
Given: sin θ = \(\frac{12}{13}\) ∴ \(\frac{P}{H}\) = \(\frac{12}{13}\)
Let, P = 12K, H = 13K
P
2
+ B
2
= H
2
…[Pythagoras’ theorem
(12K)
2
+ B
2
= (13K)
2
144K
2
+ B
2
– 169K
2
B
2
– 169 K
2
– 144K
2
= 25K
2
B = 5K
Question 17.
The average score of boys in the examination of a school is 71 and that of the girls is 73. The average score of the school in the examination is 71.8. Find the ratio of number of boys to the number of girls who appeared in the examination.
Answer:
Let the number of boys = n
1
and number of girls = n
2
Average boys’ score = 71 = \(\overline{\mathrm{X}}_1\)
Average girls’ score = 73 = \(\overline{\mathrm{X}}_2\)
Average score of school = 71.8 = \(\overline{\mathrm{X}}_{1,2}\)
Combined mean, \(\overline{\mathrm{X}}_{1,2}\) =\(\frac{n_1 \overline{\mathrm{X}}_1+n_2 \overline{\mathrm{X}}_2}{n_1+n_2}\)
71.8 = \(\frac{n_1(71)+n_2(73)}{n_1+n_2}\)
71n
1
+ 73n
2
= 71.8n
1
+ 71.8n
2
71n
1
– 71.8n
1
= 71.8n
1
– 73n
2
– 0.81n
1
= -1.2n
2
\(\frac{n_1}{n_2}\) = \(\frac{1.2}{0.8}\) ⇒ \(\frac{n_1}{n_2}\) \(\frac{3}{2}\) ⇒ n
1
: n
2
= 3 : 2
∴ No. of boys : No. of girls = 3 : 2
Question 18.
Find the other coordinates of ∆ABC with A(1, -4) and mid-points of sides through A being (2, -1) and (0, -1).
Answer:
Let B(x, y) and C(z, t) be the coordinates P is the mid-point of AB
\(\frac{x+1}{2}\) = 2
x + 1 = 4
x = 4 – 1 = 3
\(\frac{y-4}{2}\) = -1
y – 4 = -2
y = -2 + 4 = 2
∴ B(3, 2)
Q is the mid point of AC
0 = \(\frac{1+z}{2}\)
-1 = z
-1 = \(\frac{-4+t}{2}\)
-2 = -4 + t
2 = t
∴ C(-1, 2)
∴ Other coordinates of AABC are B(3, 2) and C(-1, 2).
Question 19.
Find that non-zero value of k, for which the quadratic equation kx
2
+ 1 – 2(k – 1)x + x
2
= 0 has equal roots. Hence find the roots of the equation.
Answer:
The given quadratic equation can be written as
kx
2
+ x
2
– 2(k – 1)x + 1 = 0
(k + 1) x
2
– 2 (k – 1) x + 1 = 0 ………. (i)
a = (k + 1), b = -2(k – 1), c = 1
For equal roots, D = 0
D = b
2
– 4ac
0 = [-2(k – 1)]
2
– 4 × (k + 1) × 1
0 = 4(k – 1)
2
– 4(k + 1)
0 = 4k
2
+ 4 – 8k – 4k – 4
0 = 4k
2
– 12k ⇒ 4k(k – 3) = 0
k – 3 = 0 or 4k = 0
⇒ k = 3 or k = 0 ∴ k = 3
Putting fc = 3 in equation (i), we get
4x
2
– 4x + 1 = 0
4x
2
– 2x – 2x + 1 = 0
2x(2x – 1) – 1(2x – 1) = 0
(2x – 1) (2x – 1) = 0
2x – 1 = 0 or 2x – 1 = 0
x = \(\frac{1}{2}\) or x = \(\frac{1}{2}\)
Roots are \(\frac{1}{2}\), \(\frac{1}{2}\).
Question 20.
Two different dice are rolled together. Find the probability of getting:
(i) the sum of numbers on two dice to be 5.
(ii) even numbers on both dice.
Answer:
Two dice are rolled.
∴ Total possible outcomes = 6
n
= 6
2
= 36
(i) The possible outcomes of getting a total of 5 are (2, 3), (3, 2), (1, 4), (4,1), i.e., 4.
∴ Required Probability, P(E) = \(\frac{4}{36}\) = \(\frac{1}{9}\)
(ii) The possible outcomes of getting even numbers on both dice are (2, 2), (2, 4), (2, 6), (4, 2), (4, 4), (4, 6), (6, 2), (6, 4), (6, 6) i.e., 9
∴ Required Probability, P(E) = \(\frac{9}{36}\) = \(\frac{1}{4}\)
Question 21.
In Fig. APB and AQO are semicircles, and AO = OB. If the perimeter A of the figure is 40 cm, find the area of the shaded region. [Use π = 22/7]
Answer:
OA = OB …[Given
Let OA = OB = r
Since, AO is the diameter of small semicircle
∴ Radius of small
semicircle, R = \(\frac{\mathrm{AO}}{2}\) = \(\frac{r}{2}\)
BO is the radius of big semicircle = r
A.T.Q., Total perimeter = Perimeter of small semi-circle + Perimeter of big semicircle + OB
40 = πR + πr + r
40 = \(\frac{22}{7}\)(\(\frac{r}{2}\) + \(\frac{r}{1}\)) + r \(\frac{22 r}{14}\) + \(\frac{22 r}{7}\) + \(\frac{r}{1}\)
40 = \(\frac{22 r+44 r+14 r}{14}\)
560 = 80r ⇒ r = \(\frac{560}{80}\) = 7 cm
∴ Area of shaded region = (Area of small semicircle) + (Area of big semicircle)
= \(\frac{\pi R^2}{2}\) + \(\frac{\pi r^2}{2}\) = \(\frac{\pi}{2}\)(R + r
2
)
= \(\frac{1}{2}\) × \(\frac{22}{7}\) × ((\(\frac{r}{2}\))
2
+ r
2
) = \(\frac{1}{2}\) × \(\frac{22}{7}\) ((\(\frac{7}{2}\))
2
+ (7)
2
)
= \(\frac{11}{7}\) × (\(\frac{49}{4}\) + \(\frac{49}{1}\)) = \(\frac{11}{7}\) × \(\frac{245}{4}\) = \(\frac{385}{4}\) = 96\(\frac{1}{4}\) cm
2
Question 22.
In Fig., from a cuboidal solid metallic block, of dimensions 15 cm × 10 cm × 5 cm, a cylindrical hole of diameter 7 cm is drilled ” out. Find the surface area of the remaining block.
[Use π = \(\frac{22}{7}\) ]
Answer:
Given: Length, breadth, height of cuboidal block be 15 cm, 10 cm and 5 cm respectively.
Total surface area of solid cuboidal block
= 2 (lb + bh + lh)
= 2(lb × 10 + 10 × 5 + 15 × 5) cm
2
= 2(150 + 50 + 75) = 2(275) = 550 cm
2
Radius of cylindrical hole = r = \(\frac{7}{2}\) cm
Area of two circular bases
= πr
2
+ πr
2
= 2πr
2
2 × \(\frac{22}{7}\) × \(\frac{7}{2}\) × \(\frac{7}{2}\) = 77 cm
2
Curved Surface Area of cylinder = 2πrh
= 2 × \(\frac{22}{7}\) × \(\frac{7}{2}\) × 5 = 110 cm
2
∴ Required area= (Area of cuboidal block – Area of two circular bases + Area of cylinder) = (550 – 77 + 110) cm
2
= 583 cm
2
Section – D
Questions number 23 to 30 carry 4 marks each.
Question 23.
Not in Current Syllabus.
Answer:
Not in Current Syllabus.
Question 24.
In the given figure AD = 3 cm, AE = 5 cm, BD = 4 cm, CE = 4 cm, CF = 2 cm, BF = 2.5 cm, find the pair of parallel lines and hence their lengths.
Answer:
To lines to be parallel, the required condition is equal ratio, which is true only with EF and AB as \(\frac{\mathrm{CE}}{\mathrm{EA}}\) = \(\frac{\mathrm{CF}}{\mathrm{BF}}\)
⇒ \(\frac{4}{5}\) = \(\frac{2}{2.5}\)or\(\frac{20}{25}\) … [Converse of Thales’ theorem
For lengths of parallel lines.
In ΔCEF and ΔCAB
∠C = ∠C ….[Common
∠CEF = ∠CAB …[corresponding angles
∴ ΔCEF ~ ΔCAB ..[AA similarity
⇒ \(\frac{\mathrm{CE}}{\mathrm{AC}}\) = \(\frac{\mathrm{EF}}{\mathrm{AB}}\) ⇒ \(\frac{\mathrm{CE}}{\mathrm{AE}+\mathrm{EC}}\) = \(\frac{E F}{A D+D B}\)
⇒ \(\frac{4}{9}\) = \(\frac{E F}{7}\) ⇒ EF = \(\frac{28}{9}\)
and AB = AD + DB
AB = 4 + 3 = 7
∴ EF = \(\frac{28}{9}\) cm and AB = 7 cm
Question 25.
Prove that:
(1 + cot A + tan A).(sin A – cos A) = \(\frac{\sec ^3 A-{cosec}^3 A}{\sec ^2 A \cdot {cosec}^2 A}\)
Answer:
LHS = (1 + cot A + tan A) (sin A – cos A)
Question 26.
In a class test, marks obtained by 120 students’ are given in the following frequency distri¬bution. If it is given that mean is 59, find the missing frequencies x and y.
Answer:
Σf
i
= 120 = 90 + x + y …[Given
∴ 90 + x + y = 120
x = 120 – 90 – y =30 – y …..(i)
Mean = A + \(\frac{\boldsymbol{\Sigma} f_i d_i^{\prime}}{\boldsymbol{\Sigma} f_i}\) × h
[A = 55,h = 10, Σf
i
= 120, Mean = 59
59 = 55 + (\(\frac{44+4 y}{120}\) × 10)
59 – 55 = \(\frac{4(11+y)}{12}\)
4 × 3 = 11 + y ⇒ y = 12 – 11 = 1
From (i), x = 30 – 1 = 29 ‘
∴ x = 29, y = 1
Question 27.
The numerator of a fraction is 3 less than its denominator. If 2 is added to both the numerator and the denominator, then the sum of the new fraction and original fraction is \(\frac{29}{20}\) Find the original fraction.
Or
Ramkali required ₹2500 after 12 weeks to send her daughter to school. She saved ₹100 in the first week and increased her weekly saving by ₹20 every week. Find whether she will be able to send her daughter to school after 12 weeks.
Answer:
Let the denominator and numerator of the fraction be x and x – 3 respectively.
Let the fraction be \(\frac{x-3}{x}\)
By the given condition,
⇒20[(x – 3) (x + 2) + x(x – 1)] = 29(x
2
+ 2x)
⇒ 20 (x
2
– x – 6 + x
2
– x) = 29x
2
+ 58x
⇒ 20(2x
2
– 2x – 6) = 29x
2
+ 58x
⇒ 40x
2
– 29x
2
– 40x – 58x = 120
⇒ 11x
2
– 98x – 120 = 0
⇒ 11x
2
– 110x + 12x – 120 = 0
⇒ 11x(x – 10) + 12(x – 10) = 0
⇒ (11x + 12) (x – 10) = 0
⇒ 11x + 12 = 0 or x – 10 = 0
⇒ x = \(\frac{-12}{11}\) (Reject) or x = 10
Now, denominator (x) = 10
then, numerator = x – 3 = 7
∴ The fraction is \(\frac{7}{10}\).
Or
Money required for Ramkali for admission of her daughter = ₹2500
A.P. formed by saving
100, 120, 140,… upto 12 terms …(i)
Let, a, d and n be the first term, common difference and number of terms respectively.
∴ a = 100, d = 20, n = 12
S
n
= \(\frac{n}{2}\) (2a + (n – 1 )d)
⇒ S
12
= \(\frac{12}{2}\) (2(100) + (12 – 1)20)
S
12
= \(\frac{12}{2}\) [2(100) + 11(20)] = 6[420] = ₹2520
∴ She can send her daughter to school.
Question 28.
In Fig., tangents PQ and PR are drawn from an external point P to a circle with centre O, such that ∠RPQ = 30°. A chord RS is drawn parallel to the tangent PQ. Find ∠RQS.
Answer:
Given: Join OR,
OQ and OP.
PR = PQ …[Tangents drawn from external point are equal
⇒ ∠PRQ = ∠PQR
…[In ΔPQR, Angles opposite equal sides are equal
In ΔPQR,
⇒ ∠PRQ + ∠RPQ + ∠PQR = 180° …[Δ Rule
⇒ 30 + 2∠PQR = 180°
⇒ ∠PQR = \(\frac{(180-30)^{\circ}}{2}\) = 75°
⇒ SR || QP and QR is a transversal.
∵ ∠SRQ = ∠PQR …[Alternate interior angles
∴ ∠SRQ = 75°
⇒ ∠ORP = 90° …[Tangents are ⊥ to the radius
∠OQP = 90° through the point of contact
In Quadrilateral PQOR,
∴ ∠PQO + ∠QOR + ∠ORP + ∠RPQ = 360°
90° + ∠QOR + 90° + ∠30° = 360°
∠QOR = 360° – 210° = 150°
⇒ ∠QSR = \(\frac{1}{2}\)∠QOR ⇒ ∠QSR = \(\frac{150^{\circ}}{2}\) = 75°
In ΔRSQ,
∠RSQ + ∠QRS + ∠RQS = 180° …[Δ Rule
75° + 75° + ∠RQS = 180°
∴ ∠RQS = 180° – 150° = 30°
Question 29.
From a point P on the ground the angle of elevation of the top of a tower is 30° and that of the top of a flag staff fixed on the top of the tower, is 60°. If the length of the flag staff is 5m, find the height of the tower.
Answer:
Let DC = x be the tower and AD = 5 m be the flagstaff, P is the given point so that PC = y.
In ΔPDC,
\(\frac{x}{y}\) = tan 30°= \(\frac{1}{\sqrt{3}}\)
y= √3x …….. (i)
In ΔAPC,
\(\frac{x+5}{y}\) = tan 60° = √3
\(\frac{x+5}{\sqrt{3} x}\) = √3
⇒ 3x = x + 5 or x = 2.5
∴ Height of Tower = x = 2.5 m
Question 30.
Not in Current Syllabus.
Answer:
Not in Current Syllabus.
