Class 10 Some Applications of Trigonometry Mock Test (CBSE)

Topics in Some Applications of Trigonometry

• Heights and distances
• Angles of elevation and depression

Some Applications of Trigonometry practice questions with answers

1. Q1Multiple choice·1 mark
   A ladder $15\text{ m}$ long just reaches the top of a vertical wall. If the ladder makes an angle of $60^\circ$ with the ground, then the height of the wall is:

   1. (a)$\frac{15\sqrt{3}}{2}\text{ m}$
   2. (b)$15\sqrt{3}\text{ m}$
   3. (c)$\frac{15}{2}\text{ m}$
   4. (d)$15\text{ m}$
   Show answer and step marks

   1. Correct option: (a) $\frac{15\sqrt{3}}{2}\text{ m}$. Height = $15 \sin 60^\circ = 15 \times \frac{\sqrt{3}}{2} = \frac{15\sqrt{3}}{2}\text{ m}$+1

   Answer: (a)
2. Q2Multiple choice·1 mark
   An aeroplane flying horizontally at a uniform speed, $1000\sqrt{3}\text{ m}$ above the ground, is moving directly away from an observer on the ground. Its angle of elevation from the observer is $60^\circ$, and after $10\text{ seconds}$ it is $30^\circ$. The speed of the aeroplane is:

   1. (a)$720\text{ km/h}$
   2. (b)$360\text{ km/h}$
   3. (c)$240\text{ km/h}$
   4. (d)$1440\text{ km/h}$
   Show answer and step marks

   1. Correct option: (a) $720\text{ km/h}$. Calculating the horizontal distances for both angles of elevation gives $1000\text{ m}$ and $3000\text{ m}$. The distance travelled in $10\text{ s}$ is $2000\text{ m}$, giving a speed of $200\text{ m/s}$ or $720\text{ km/h}$.+1

   Answer: (a)
3. Q3Multiple choice·1 mark
   A ladder $10\text{ m}$ long reaches a window $5\text{ m}$ above the ground. The angle of elevation of the ladder from the ground is:

   1. (a)$60^\circ$
   2. (b)$30^\circ$
   3. (c)$45^\circ$
   4. (d)$90^\circ$
   Show answer and step marks

   1. Correct option: (b) $30^\circ$. Let $\theta$ be the angle of elevation. Then $\sin \theta = \frac{5}{10} = \frac{1}{2}$, which gives $\theta = 30^\circ$.+1

   Answer: (b)
4. Q4Short answer·2 marks
   A kite is flown on a windy day. A child holds the string at a point $1.5$ m above the ground. The string is straight and taut, with length $40$ m from the child's hand to the kite, and the angle of elevation of the kite as seen from the child's hand is $30^\circ$. (a) Find the vertical height of the kite above the child's hand. [1] (b) Explain why the height of the child's hand above the ground must be added to the answer in (a) to find the height of the kite above the ground, and state that height. [1]
   Show answer and step marks

   1. (a) Let the height of the kite above the hand be $h$ m. Using $\sin 30^\circ = \dfrac{h}{40}$+0.5
   2. $h = 40 \times \dfrac{1}{2} = 20$ m+0.5
   3. (b) The angle of elevation is measured from the hand, so $20$ m is the rise above the hand; the hand is $1.5$ m above the ground, so the kite's height above the ground is $20 + 1.5 = 21.5$ m+1

   Answer: (a) 20 m (b) The rise is measured from the hand, which is 1.5 m above ground, so add 1.5 m; height above ground = 21.5 m.
5. Q5Short answer·3 marks
   From the top $T$ of a lighthouse, $36$ m above the sea level, the angle of depression of a fishing boat $B$ on the sea is $30^\circ$. The boat and the foot $L$ of the lighthouse lie on a horizontal line at sea level, on the same side of the lighthouse. Take $\sqrt{3} = 1.73$. (a) Find the distance of the boat from the foot of the lighthouse. (b) The boat then sails straight towards the lighthouse, and from $T$ the angle of depression of the boat becomes $60^\circ$. Find the distance the boat has sailed.
   Show answer and step marks

   1. (a) Angle of elevation of $T$ from $B$ equals the angle of depression $30^\circ$ (alternate angles), so $\tan 30^\circ = \dfrac{36}{LB}$+1
   2. $LB = 36\sqrt{3} \approx 62.28$ m+0.5
   3. (b) For the new position $B'$, $\tan 60^\circ = \dfrac{36}{LB'}$+0.5
   4. $LB' = \dfrac{36}{\sqrt{3}} = 12\sqrt{3} \approx 20.76$ m+0.5
   5. Distance sailed $= BB' = LB - LB' = 36\sqrt{3} - 12\sqrt{3} = 24\sqrt{3} \approx 41.52$ m+0.5

   Answer: (a) $36\sqrt{3} \approx 62.28$ m; (b) $24\sqrt{3} \approx 41.52$ m
6. Q6Short answer·3 marks
   As observed from the top of a $75\text{ m}$ high lighthouse from the sea-level, the angles of depression of two ships are $30^\circ$ and $45^\circ$ if one ship is exactly behind the other on the same side of the lighthouse. Find the distance between the two ships. (Use $\sqrt{3} = 1.732$)
   Show answer and step marks

   1. Let $AB = 75\text{ m}$ be the lighthouse, $C$ the nearer ship ($45^\circ$) and $D$ the farther ($30^\circ$). In right $\triangle ABC$: $BC = \frac{AB}{\tan 45^\circ} = 75\text{ m}$.+1
   2. In right $\triangle ABD$: $BD = \frac{AB}{\tan 30^\circ} = 75\sqrt{3}\text{ m}$.+1
   3. $CD = BD - BC = 75\sqrt{3} - 75 = 75(\sqrt{3} - 1)$.+0.5
   4. $CD = 75(1.732 - 1) = 75 \times 0.732 = 54.9\text{ m}$.+0.5

   Answer: $54.9\text{ m}$

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