CBSE 10th Standard Maths Subject Some Applications of Trigonometry Case Study Questions With Solution 2021
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CBSE 10th Standard Maths Subject Some Applications of Trigonometry Case Study Questions With Solution 2021
10th Standard CBSE
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Reg.No. :
Maths
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There are two temples on each bank of a river. One temple is 50 m high. A man, who is standing on the top of 50 m high temple, observed from the top that angle of depression of the top and foot of other temple are 30° and 60° respectively. (Take \(\sqrt{3}\) = 1.73)
Based on the above information, answer the following questions.
(i) Measure of \(\angle\)ADF is equal to(a) 45° (b) 60° (c) 30° (d) 90° (ii) Measure of \(\angle\)ACB is equal to
(a) 45° (b) 60° (c) 30° (d) 90° (iii) Width of the river is
(a) 28.90 m (b) 26.75 m (c) 25 m (d) 27 m (iv) Height of the other temple is
(a) 32.5 m (b) 35 m (c) 33.33 m (d) 40 m (v) Angle of depression is always
(a) reflex angle (b) straight (c) an obtuse angle (d) an acute angle (a) -
A circus artist is climbing through a 15 m long rope which is highly stretched and tied from the top of a vertical pole to the ground as shown below. Based on the above information, answer the following questions.
(i) Find the height of the pole, if angle made by rope to the ground level is 45°.\((a) 15 \mathrm{~m}\) \((b) 15 \sqrt{2} \mathrm{~m}\) \((c) \frac{15}{\sqrt{3}} \mathrm{~m}\) \((d) \frac{15}{\sqrt{2}} \mathrm{~m}\) (ii) If the angle made by the rope to the ground level is 45°, then find the distance between artist and pole at ground level.
\((a) \frac{15}{\sqrt{2}} \mathrm{~m}\) \((b) 15 \sqrt{2} \mathrm{~m}\) \((c) 15 \mathrm{~m}\) \((d) {15}{\sqrt{3}} \mathrm{~m}\) (iii) Find the height of the pole if the angle made by the rope to the ground level is 30°.
(a) 2.5 m (b) 5 m (c) 7.5 m (d) 10 m (iv) If the angle made by the rope to the ground level is 30° and 3 m rope is broken, then find the height of the pole
(a) 2m (b) 4m (c) 5m (d) 6m (v) Which mathematical concept is used here?
(a) Similar Triangles (b) Pythagoras Theorem (c) Application of Trigonometry (d) None of these (a) -
An electrician has to repair an electric fault on the pole of height of8 m. He needs to reach a point 2 m below the top of the pole to undertake the repair work.
Based on the above information, answer the following questions.
(i) Length of BD is(a) 10 m (b) 6 m (c) 4 m (d) 4 m (ii) What should be the length of ladder, so that it makes an angle of 60° with the ground?
\((a) 4\sqrt{3} {~m}\) \((b) 2\sqrt{3} {~m}\) \((c) 3\sqrt{3} {~m}\) \((d) 5\sqrt{3} {~m}\) (iii) The distance between the foot ofladder and pole is
\((a) 6\sqrt{3} {~m}\) \((b) 4\sqrt{3} {~m}\) \((c) 3\sqrt{3} {~m}\) \((d) 2\sqrt{3} {~m}\) (iv) What will be the measure of \(\angle\)BCD when BD and CD are equal?
(a) 30° (b) 45° (c) 60° (d) 75° (v) Find the measure of \(\angle\)DBC.
(a) 15° (b) 60° (c) 30° (d) 45° (a) -
A boy is standing on the top of light house. He observed that boat P and boat Q are approaching to light house from opposite directions. He finds that angle of depression of boat P is 45° and angle of depression of boat Q is 30°. He also knows that height of the light house is 100 m.
Based on the above information, answer the following questions.
(i) Measure of \(\angle\)ACD is equal to(a) 30° (b) 45° (c) 60° (d) 90° (ii) If \(\angle\)YAB = 30°, then \(\angle\)ABD is also 30°, Why?
(a) vertically opposite angles (b) alternate interior angles (c) alternate exterior angles (d) corresponding angles (iii) Length of CD is equal to
(a) 90 m (b) 60 m (c) 100 m (d) 80 m (iv) Length of BD is equal to
(a) 50 m (b) 100 m (c) 100\(\sqrt{2}\) m (d) 100\(\sqrt{3}\) m (v) Length of AC is equal to
(a)100\(\sqrt{2}\) m (b) 100\(\sqrt{3}\) m (c) 50 m (d) 100 m (a) -
Teewan, Arun and Pankaj were celebrating the festival of Diwali in open ground with firecrackers. There is a pedestal in the ground. All of sudden Teewan stands on pedestal and release sky lantern from the top of pedestal.
Based on the above information answer the following questions. (Take \(\sqrt{3}\) = J .73)
(i) Which one is a pair of angle of depression?\((a) (\angle x, \angle y)\) \((b) (\angle y, \angle z)\) \((c) (\angle z, \angle t)\) \((d) (\angle r, \angle q)\) (ii) If the position of Pankaj is 25 m away from the base of pedestal and Zr = 30°, then find the height of pedestal.
(a) 14.45m (b) 15.5m (c) 16.36m (d) 17.36m (iii) If the height of pedestal is 30 m, \(\angle\)t = 45° and \(\angle\)z = 30°, then the horizontal distance between Arun and Pankaj is
(a) 24.5 m (b) 19.5 m (c) 20 m (d) 21.9 m (iv) If the vertical height of sky lantern from the top of pedestal is 12 m and \(\angle\)y = 30°, then distance between Teewan and sky lantern is
(a) 20 m (b) 16.97 m (c) 24 m (d) 19.86 m (v) If \(\angle\)q = 60° and position of Arun is 15 m away from the base of pedestal, then find the height of pedestal.
(a) 16.25 m (b) 25 m (c) 25.95 m (d) 26 m (a)
Case Study Questions
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CBSE 10th Standard Maths Subject Some Applications of Trigonometry Case Study Questions With Solution 2021 Answer Keys
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(i) (c) : Since AE || FD
\(\therefore\) \(\angle\)EAD = \(\angle\)ADF = 30° [Alternate interior angles]
(ii) (b): Since, AE || BC
\(\therefore\) \(\angle\)EAC = \(\angle\)ACB = 60° [Alternate interior angles]
(iii) (a) : In \(\Delta\)ABC,
\(\begin{array}{l} \tan 60^{\circ}=\frac{A B}{B C} \Rightarrow \sqrt{3}=\frac{50}{B C} \\ \Rightarrow \quad B C=\frac{50}{\sqrt{3}}=28.90 \mathrm{~m} \end{array}\)
(iv) (c): In \(\Delta\)ADF, \(\tan 30^{\circ}=\frac{A F}{F D}\)
\(\Rightarrow \frac{1}{\sqrt{3}}=\frac{A B-B F}{F D} \Rightarrow \frac{1}{\sqrt{3}}=\frac{50-C D}{\frac{50}{\sqrt{3}}}\)
\(\left[\because F D=B C=\frac{50}{\sqrt{3}}\right] \)
\(\Rightarrow \frac{50}{3}=50-C D \Rightarrow C D=50-\frac{50}{3}=\frac{100}{3}=33.33 \mathrm{~m}\)
(v) (d) -
(i) (d): Let h be the height of the pole.
In \(\Delta\)ABC,
\( \frac{h}{15}=\sin 45^{\circ} \Rightarrow \frac{h}{15}=\frac{1}{\sqrt{2}}\)
\(\Rightarrow \quad h =\frac{15}{\sqrt{2}} \mathrm{~m}\)
(ii) (a): Let x be the required distance.
In \(\Delta\)ABC,
\(\frac{x}{15}=\cos 45^{\circ}=\frac{1}{\sqrt{2}}\)
\(\Rightarrow \quad x=\frac{15}{\sqrt{2}} \mathrm{~m}\)
(iii) (c) : Let h be the height of the pole.
In right triangle ABC,
\(\frac{h}{15}=\sin 30^{\circ}=\frac{1}{2}\)
\(\Rightarrow \quad h=\frac{15}{2}=7.5 \mathrm{~m}\)
(iv) (d): If 3 m rope is broken, then the length of the rope is 12 m.
\(\text { In } \Delta A B C, \frac{h}{12}=\sin 30^{\circ}=\frac{1}{2} \)
\(\Rightarrow \quad h=\frac{12}{2}=6 \mathrm{~m}\)
(v) (c) -
(i) (b): Total height of pole = 8 m
\(\therefore\) BD = AD - AB = (8 - 2)m = 6 m
(ii) (a): \(\text { In } \Delta B D C, \frac{B D}{B C}=\sin 60^{\circ}\)
\(\Rightarrow \quad \frac{6}{B C}=\frac{\sqrt{3}}{2} \)
\(\Rightarrow \quad B C=\frac{12}{\sqrt{3}} \times \frac{\sqrt{3}}{\sqrt{3}}=4 \sqrt{3} \mathrm{~m}\)
(iii) (d): \(\text { In } \triangle B D C\)
\(\frac{B D}{C D}=\tan 60^{\circ} \Rightarrow \frac{6}{C D}=\sqrt{3} \Rightarrow C D=\frac{6}{\sqrt{3}} \times \frac{\sqrt{3}}{\sqrt{3}}=2 \sqrt{3} \mathrm{~m}\)
(iv) (b) : \(\text { If } \Delta B C D\)
\(\frac{B D}{C D}=\tan \theta \Rightarrow 1=\tan \theta \quad[\because B D=C D] \)
\(\Rightarrow \quad \theta=45^{\circ}\)
(v) (c) : \(\operatorname{In} \Delta B D C, \angle B+\angle D+\angle C=180^{\circ}\)
\(\therefore \quad \angle B=180^{\circ}-60^{\circ}-90^{\circ}=30^{\circ}\) -
(i) (b): \(\angle X A C=45^{\circ}\)
\(\therefore \quad \angle A C D=45^{\circ}\) [Alternate interior angles]
(ii) (b)
(iii) (c) : \(\text { In } \Delta A C D\)
\(\frac{A D}{D C}=\tan 45^{\circ} \)
\(\Rightarrow \frac{100}{D C}=1 \Rightarrow D C=100 \mathrm{~m}\)
(iv) (d): \(\text { In } \Delta A B D, \frac{A D}{B D}=\tan 30^{\circ}\)
\(\Rightarrow \quad \frac{100}{B D}=\frac{1}{\sqrt{3}} \)
\(\Rightarrow \quad B D=100 \sqrt{3} \mathrm{~m}\)
(v) (a): \(\text { In } \Delta A D C\)
\(\frac{A D}{A C}=\sin 45^{\circ} \Rightarrow \frac{100}{A C}=\frac{1}{\sqrt{2}} \Rightarrow A C=100 \sqrt{2} \mathrm{~m}\) -
(i) (c)
(ii) (a): Let AB be the height of pedestal.
\(\text { In } \Delta A B C, \)
\(\tan 30^{\circ}=\frac{A B}{B C} \)
\(\Rightarrow \quad A B=\frac{25}{\sqrt{3}}=\frac{25}{1.73}=14.45 \mathrm{~m}\)
(iii) (d): Let x be the distance between Arun and Pankaj.
\(\text { In } \Delta A B \dot{D}, \tan 45^{\circ}=\frac{A B}{B D}\)
\(\Rightarrow B D=30 \mathrm{~m}\)
\(\text { Now, in } \Delta \overline{A B C} \text { , }\)
\(\tan 30^{\circ}=\frac{A B}{B C}\)
\(\Rightarrow \frac{30}{30+x}=\frac{1}{\sqrt{3}}\)
\(\Rightarrow x=30(\sqrt{3}-1)=30 \times 0.73=21.9 \mathrm{~m}\)
(iv) (c): \(\text { In } \triangle A R S\)
\(\sin 30^{\circ}=\frac{R S}{A S} \)
\(\Rightarrow \quad \frac{12}{A S} =\frac{1}{2} \Rightarrow A S=12 \times 2=24 \mathrm{~m}\)
(v) (c): \(\text { In } \Delta A B D, \frac{A B}{B D}=\tan 60^{\circ}\)
\(\begin{array}{l} \Rightarrow \frac{A B}{15}=\sqrt{3} \\ \Rightarrow A B=15 \times 1.73=25.95 \mathrm{~m} \end{array}\)
Case Study Questions