{"id":37210,"date":"2022-02-07T18:10:59","date_gmt":"2022-02-07T12:40:59","guid":{"rendered":"https:\/\/mcq-questions.com\/?p=37210"},"modified":"2022-02-25T09:38:59","modified_gmt":"2022-02-25T04:08:59","slug":"mcq-questions-for-class-10-maths-chapter-9","status":"publish","type":"post","link":"https:\/\/mcq-questions.com\/mcq-questions-for-class-10-maths-chapter-9\/","title":{"rendered":"MCQ Questions for Class 10 Maths Chapter 9 Some Applications of Trigonometry"},"content":{"rendered":"
Question 1.<\/p>\n
(A) 60\u00b0
\n(B) 45\u00b0
\n(C) 30\u00b0
\n(D) 90\u00b0
\nAnswer:
\n(A) 60\u00b0<\/p>\n
Explanation:
\nIn \u2206ABC, \u2220B = 90\u00b0
\ntan\u03b8 = \\(\\sqrt{3}\\) = \u221a3 = tan 60\u00b0 \u21d2 \u03b8 = 60\u00b0<\/p>\n
<\/p>\n
Question 2.<\/p>\n
(A) 50 \u221a3
\n(B) 150 \u221a3
\n(C) 150 \u221a2
\n(D) 75
\nAnswer:
\n(B) 150 \u221a3<\/p>\n
Explanation:
\n\u2206ABC, \u2220B = 90\u00b0
\ntan 30\u00b0 = \\(\\frac{150}{x}\\)
\n\\(\\frac{1}{\\sqrt{3}}\\) = \\(\\frac{150}{x}\\)
\nx = 150 \u221a3 m
\n<\/p>\n
Question 3.<\/p>\n
(A) 75 m
\n(B) 79.41 m
\n(C) 80 m
\n(D) 72.5 m
\nAnswer:
\n(A) 75 m<\/p>\n
Explanation:
\ntan \u03b8 = \\(\\frac{15}{8}\\)
\nsin \u03b8 = \\(\\frac{15}{17}\\)
\nNow, sin \u03b8 = \\(\\frac{x}{85}\\)
\nFrom, equation (i) and (ii),
\n\\(\\frac{15}{17}\\) = \\(\\frac{x}{85}\\)
\nx = 75 m
\n<\/p>\n
Question 4.<\/p>\n
(A) 30\u00b0
\n(B) 60\u00b0
\n(C) 45\u00b0
\n(D) 75\u00b0
\nAnswer:
\n(B) 60\u00b0<\/p>\n
Explanation:
\nLet the length of shadow is x, Then height of pale = \u221a3 x
\ntan \u03b8 = \\(\\frac{C B}{A B}\\)
\ntan \u03b8 = \\(\\frac{\\sqrt{3} x}{x}\\)
\ntan \u03b8 = \u221a3
\ntan \u03b8 = tan 60\u00b0
\n\u03b8 = 60\u00b0
\n<\/p>\n
<\/p>\n
Question 5.<\/p>\n
(A) 25 \u221a3
\n(B) 50 \u221a3
\n(C) 75 \u221a3
\n(D) 150
\nAnswer:
\n(C) 75 \u221a3<\/p>\n
Explanation:
\nIn \u2206ABC, \u2220B = 90\u00b0
\ntan \u03b8 = \\(\\frac{C B}{A B}\\)
\ntan 30\u00b0 = \\(\\frac{75}{x}\\)
\n\\(\\frac{1}{\\sqrt{3}}\\) = \\(\\frac{75}{x}\\)
\nx = 75 \u221a3 m<\/p>\n
<\/p>\n
Question 6.<\/p>\n
(A) 20 m
\n(B) 40 m
\n(C) 60 m
\n(D) 80 m
\nAnswer:
\n(B) 40 m<\/p>\n
Explanation:
\nIn \u2206ABC, \u221aA = 90\u00b0
\ntan \u03b8 = \\(\\frac{C A}{A B}\\)
\ntan \u03b8 = \\(\\frac{20}{x}\\)
\nx = \\(\\frac{20}{\\tan \\theta}\\) ….. (i)
\n
\nIn \u2206CDE, \u221aD = 90\u00b0
\nNow, tan 0 = \\(\\frac{E D}{C D}\\)
\ntan 0 = \\(\\frac{h-20}{x}\\)
\nx = \\(\\frac{h-20}{\\tan \\theta}\\) …….(ii)
\nfrom equation (i) and (ii),
\n\\(\\frac{h-20}{\\tan \\theta}\\) = \\(\\frac{20}{\\tan \\theta}\\)
\nh – 20 = 20
\nh = 40 m<\/p>\n
Assertion and Reason Based MCQs<\/span><\/p>\n Directions: In the following questions, A statement of Assertion (A) is followed by a statement of Reason (R). Mark the correct choice as. Question 1.<\/p>\n Answer: Explanation: <\/p>\n Question 2.<\/p>\n Answer: Explanation: Question 3.<\/p>\n Answer: Explanation: Case – Based MCQs<\/span><\/p>\n Attempt any four sub-parts from each question. Each sub-part carries 1 mark. Question 1.<\/p>\n (A) 1.3 m Explanation: <\/p>\n Question 2.<\/p>\n (A) 4.28 m Explanation: Question 3.<\/p>\n (A) 3 . 7 Explanation: Question 4.<\/p>\n (A) 7.4 m Explanation: <\/p>\n Question 5.<\/p>\n (A) 60\u00b0 Explanation: II. Read the following text and Answer: the following question on the basis of the same: Question 1.<\/p>\n (A) 30\u00b0 Explanation: <\/p>\n Question 2.<\/p>\n (A) 25.24 m Explanation: Question 3.<\/p>\n (A) 20 \u221a3 m Explanation: Question 4.<\/p>\n (A) 30\u00b0 <\/p>\n Question 5.<\/p>\n (A) corresponding angle III. Read the following text and Answer: the following question on the basis of the same: Question 1.<\/p>\n (A) 1118.36 km Explanation: <\/p>\n Question 2.<\/p>\n (A) 1139.4 km Explanation: Question 3.<\/p>\n (A) 1139.4 km Question 4.<\/p>\n (A) 30\u00b0 Explanation: <\/p>\n Question 5.<\/p>\n (A) 1118.327 km Some Applications of Trigonometry Class 10 MCQ Questions with Answers Question 1. A pole 6 m high casts a shadow 2\u221a3 m long on the ground, then the Sun’s elevation is : (A) 60\u00b0 (B) 45\u00b0 (C) 30\u00b0 (D) 90\u00b0 Answer: (A) 60\u00b0 Explanation: In \u2206ABC, \u2220B = 90\u00b0 tan\u03b8 = = \u221a3 = tan …<\/p>\n
\n(A) Both A and R are true and R is the correct explanation of A
\n(B) Both A and R are true but R is NOT the correct explanation of A
\n(C) A is true but R is false
\n(D) A is false and R is True<\/p>\nAssertion (A): If the ratio of the length of a vertical rod and the length of its shadow is 1 : \u221a3, then the angle of elevation of the sun at that moment is 30\u00b0.
\nReason (R): If the ratio of the height of a tower and the length of its shadow on the ground is \u221a3 : 1, then the angle of elevation of the sun is 60\u00b0.<\/h2>\n
\n(B) Both A and R are true but R is NOT the correct explanation of A<\/p>\n
\nIn case of assertion:
\nLet AB be a vertical rod and BC be its shadow. From the figure, \u2220ACB = \u03b8
\nIn \u2206ABC,
\n
\ntan \u03b8 = \\(\\frac{A B}{B C}\\)
\ntan \u03b8 = \\(\\frac{1}{\\sqrt{3}}\\)
\n\u2234 \\(\\frac{A B}{B C}\\)=\\(\\frac{1}{\\sqrt{3}}\\)
\ntan = tan 30\u00b0
\n\u03b8 = 30\u00b0
\n\u2234Assertion is correct.
\nIN case of reason:
\n
\nLEt the height of tower be AB and its shadow be BC.
\n\u2234\\(\\frac{B C}{A B}\\) = tan \u03b8
\n= \\(\\frac{\\sqrt{3}}{1}\\)
\n= tan 60\u00b0
\nHence, the angle of elevation of are correct but reason iss not the correct explanation for assertion.<\/p>\nAssertion (A): A ladder 15 m long lust reaches the top of a vertical wall. If the ladder makes an angle of 60r with the wall, then the height of the wall is 75 m.
\nReason (R): If the angle of elevation of the top of a tower from a point on the ground, which is 30 m away from the foot of the tower is 30\u00b0, then the height of the tower is 10 \u221a3 m.<\/h2>\n
\n(D) A is false and R is True<\/p>\n
\nIn case of assertion:
\nLet AW be the ladder and WL be the wall.
\nIn \u2206AWL, ..
\ncos 60\u00b0 = \\(\\frac{x}{15}\\)
\n\u21d2 \\(\\frac{1}{2}\\) = \\(\\frac{x}{15}\\)
\n\u21d2 x = 7 . 5
\nHence, the required height of the wall is 7.5 m.
\n\u2234Assertion is incorrect.
\nIn case of reason:
\nLet AB be the tower.
\n
\nIN \u2206ABC,
\ntan 30 = \\(\\frac{A B}{30}\\)
\n\u21d2 \\(\\frac{1}{\\sqrt{3}}\\) = \\(\\frac{A B}{30}\\)
\n\u21d2 AB = 10 \u221a3
\nHence, the required height of the tower is 10 \u221a3 m.
\n\u2234 Reason is correct.
\nHence, assertion is incorrect but reason is correct<\/p>\nAssertion (A): If the length of the ladder placed against a wall is twice the distance between the foot of the ladder and the wall, then the angle made by the ladder with the horizontal is 60\u00b0.
\nReason (R): If a tower is 20 m high and its shadow on the ground is 20 \u221a3 long, then the sun’s altitude is 60\u00b0.<\/h2>\n
\n(C) A is true but R is false<\/p>\n
\nIn case of assertion:
\nLet the distance between the foot of she ladder Iander and the wall, AB be x.
\nthen AC, the length of the laddre = 2x
\n
\nIN \u2206ABC, \u2220B = 90\u00b0
\ncos A = \\(\\frac{x}{2 x}\\)
\ncos A = \\(\\frac{1}{2 x}\\) = c0s 60\u00b0
\nA = 60\u00b0
\n\u2234Assertion is incorrect.
\nIn case of reason:
\nLet the ACB be \u03b8, \u2220B = 90\u00b0
\n
\ntan \u03b8 = \\(\\frac{A B}{B C}\\)
\ntan \u03b8 = \\(\\frac{20}{20 \\sqrt{3}}\\) = \\(\\frac{1}{\\sqrt{3}}\\) = tan 30\u00b0
\n\u21d2 \u03b8 = 30\u00b0
\nThus, assertion is correct but reason is incorrect.<\/p>\n
\nI. Read the following text and Answer: the following question on the basis of the same: An electrician has to repaired an electric fault on the pole of height 5 m. She needs to reach a point 1.3 m below the top of the pole to undertake the repair work (see figure)
\n<\/p>\nWhat is the length of BD?<\/h2>\n
\n(B) 5 m
\n(C) 3.7 m
\n(D) None of these
\nAnswer:
\n(C) 3.7 m<\/p>\n
\nFrom figure, the electrician is required to reach at the point B on the pole AD. So, BD – AD – AB
\n= (5- 1.3) m = 3.7 m<\/p>\nWhat should be the length of Ladder, when inclined at an angle of 60\u00b0 to the horizontal ?<\/h2>\n
\n(B) \\(\\frac{3.7}{\\sqrt{3}}\\)
\n(C) 3.7 m
\n(D) 7 . 4
\nAnswer:
\n(A) 4.28 m<\/p>\n
\nIn \u2206ADC,
\nsin 60 = \\(\\frac{B D}{B C}\\)
\n\\(\\frac{\\sqrt{3}}{2}\\) = \\(\\frac{3.7}{B C}\\)
\nBC = \\(\\frac{3.7 \\times 2}{\\sqrt{3}}\\)
\nBC = 4 . 28 m(approx)<\/p>\nHow far from the foot of pole should she place the foot of the ladder?<\/h2>\n
\n(B) 2 . 12
\n(C) \\(\\frac{1}{\\sqrt{3}}\\)
\n(D) None of these
\nAns.
\n(B) 2 . 12<\/p>\n
\nIn \u2206BDC,
\ncot 60 = \\(\\frac{D C}{B D}\\)
\n\u21d2 \\(\\frac{1}{\\sqrt{3}}\\) = \\(\\frac{D C}{3.7}\\)
\n\u21d2 DC = \\(\\frac{3.7}{\\sqrt{3}}\\)
\n\u21d2 DC = 2 . 14 m (approx)<\/p>\nIf the horizontal angle is changed to 30\u00b0, then what should be the length of the ladder ?<\/h2>\n
\n(B) 3.7 m
\n(C) 1.3 m
\n(D) 5 m
\nAnswer:
\n(A) 7.4 m<\/p>\n
\nIn \u2206BDC,
\n\u2234 sin 60\u00b0 = \\(\\frac{B D}{B C}\\)
\n\u21d2 \\(\\frac{1}{2}\\) = \\(\\frac{3.7}{B C}\\)
\n\u21d2 BC = 3 . 7 x 2 = (90\u00b0 + 60\u00b0)
\n= 30\u00b0<\/p>\nWhat is the value of \u2220B ?<\/h2>\n
\n(B) 90\u00b0
\n(C) 30\u00b0
\n(D) 180\u00b0
\nAnswer:
\n(C) 30\u00b0<\/p>\n
\nIn \u2206ADC, angle D is 90\u00b0. So, by angle sum property.
\n\u2220B + \u2220D + \u2220C = 180\u00b0 or, \u2220B = 180\u00b0 – (90\u00b0 + 60\u00b0)
\n= 30\u00b0<\/p>\n
\nA group of students of class X visited India Gate on an education trip. The teacher and students had interest in historv as well. The teacher narrated that India Gate, official name Delhi Memorial, originally called All-India War Memorial, monumental sandstone arch in New Delhi, dedicated to the troops of British India who died in wars fought between 1914 and 1919. The teacher also said that India Gate, which is located at the eastern end of the Raj path (formerly called the Kings way), is about 138 feet (42 metres) in height.
\n<\/p>\nWhat is the angle of elevation if they are standing at a distance of 42 m awav from the monument ?<\/h2>\n
\n(B) 45\u00b0
\n(C) 60\u00b0
\n(D) 0\u00b0
\nAnswer:
\n(B) 45\u00b0<\/p>\n
\nHeight of INdian gate = 42 m Distance between students and indian gate = 42 cm
\n
\nNow, tan \u03b8 = \\(\\frac{A B}{B C}\\)
\ntan \u03b8 = \\(\\frac{42}{42}\\)
\ntan \u03b8 = 1
\ntan \u03b8 = tan 45\u00b0
\n\u03b8 = 45\u00b0
\nHence, angle of elevation = 45\u00b0<\/p>\nThey want to see the tower at an angle of 60\u00b0. So, they want to know the distance where they should stand and hence find the distance.<\/h2>\n
\n(B) 20.12 m
\n(C) 42 m
\n(D) 24.64 m
\nAnswer:
\n(A) 25.24 m<\/p>\n
\nHeight of India gate = 42 cm \u2019Angle = 60\u00b0 Let.the distance between students and India gate = 42 m.
\n
\nNow, tan 0 = \\(\\frac{A B}{B C}\\)
\ntan 60 = \\(\\frac{42}{x}\\)
\n\u221a3 = \\(\\frac{42}{x}\\)
\nx = \\(\\frac{42}{\\sqrt{3}}\\)
\nx = \\(\\frac{42 \\times \\sqrt{3}}{\\sqrt{3} \\times \\sqrt{3}}\\)
\n= \\(\\frac{42 \\sqrt{3}}{3}\\)
\n= 14 3 m = 25. 24 m<\/p>\nIf the altitude of the Sun is at 60\u00b0, then the height of the vertical tower that will cast a shadow of length 20 m is<\/h2>\n
\n(B) \\(\\frac{20}{\\sqrt{3}}\\)
\n(C) \\(\\frac{15}{\\sqrt{3}} \\mathrm{~m}\\)
\n(D) 15 \u221a3
\nAnswer:
\n(A) 20 \u221a3 m<\/p>\n
\n
\nLet, the height of tower = h
\nNow, tan 0 = \\(\\frac{A B}{B C}\\)
\ntan 60 = \\(\\frac{h}{20}\\)
\n\u221a3 = \\(\\frac{h}{20}\\)
\nh = 20 \u221a3<\/p>\nThe ratio of the length of a rod and its shadow is 1 : 1. The angle of elevation of the Sun is<\/h2>\n
\n(B) 45\u00b0
\n(C) 60\u00b0
\n(D) 90\u00b0
\nAnswer:
\n(B) 45\u00b0<\/p>\nThe angle formed bv the line of sight with the horizontal when the object viewed is below the horizontal level is<\/h2>\n
\n(B) angle of elevation
\n(C) angle of depression
\n(D) complete angle
\nAnswer:
\n(A) corresponding angle<\/p>\n
\nA Satellite flying at height h is watching the top of the two tallest mountains in Uttarakhand and Karnataka ,them being Nanda Devi(height 7,816 m) and Mullayanagiri (height 1,930 m). The angles of depression from the satellite, to the top of Nanda Devi and Mullayanagiri are 30c and 60\u00b0 respectively. If the distance between the peaks of two mountains is 1937 km, and the satellite is vertically above the midpoint of the distance between the two mountains.
\n<\/p>\nThe distance of the satellite from the top of Nanda Devi is<\/h2>\n
\n(B) 577.52 km
\n(C) 1937 km
\n(D) 1025.36 km
\nAnswer:
\n(A) 1118.36 km<\/p>\n
\n
\nNow, AG = \\(\\frac{1937}{2}\\) km
\ncos \u03b8 = \\(\\frac{A G}{A F}\\)
\ncos \u03b8 = \\(\\frac{\\frac{1937}{2}}{A F}\\)
\n\\(\\frac{\\sqrt{3}}{2}\\) = \\(\\frac{1937}{2 A F}\\)
\nAF = \\(\\frac{1937}{\\sqrt{3}}\\)
\nAF = 1118.36 km<\/p>\nThe distance of the satellite from the top of Mullayanagiri is<\/h2>\n
\n(B) 577.52 km
\n(C) 1937 km
\n(D) 1025.36 km
\nAnswer:
\n(C) 1937 km<\/p>\n
\ncos \u03b8 = \\(\\frac{P H}{F P}\\)
\ncos 60\u00b0 = \\(\\frac{1937}{2 F P}\\)
\n\\(\\frac{1}{2}\\) = \\(\\frac{1937}{2 F P}\\)
\nFP = 1937 km<\/p>\nThe distance of the satellite from the ground is<\/h2>\n
\n(B) 577.52 km
\n(C) 1937 km
\n(D) 1025.36 km
\nAnswer:
\n(B) 577.52 km<\/p>\nWhat is the angle of elevation if a man is standing at a distance of 7816 m from Xanda Devi ?<\/h2>\n
\n(B) 45\u00b0
\n(C) 60\u00b0
\n(D) 0\u00b0
\nAnswer:
\n(B) 45\u00b0<\/p>\n
\nHeight of Nanda Devi Mountain = 7816 m
\nDistance between man and mountain = 7816 m.
\n
\ntan \u03b8 = \\(\\frac{A B}{B C}\\)
\ntan \u03b8 = \\(\\frac{7816}{7816}\\)
\ntan \u03b8 = 1
\ntan \u03b8 = tan 45\u00b0
\n\u03b8 = 45\u00b0<\/p>\nIf a mile stone very far away from, makes 45\u00b0 to the top of Mullanyangiri mountain. So, find the distance of this mile stone from the mountain.<\/h2>\n
\n(B) 566.976 km
\n(C) 1937 km
\n(D) 1025.36 km
\nAnswer:
\n(C) 1937 km<\/p>\nMCQ Questions for Class 10 Maths with Answers<\/a><\/h4>\n","protected":false},"excerpt":{"rendered":"