Some Applications of Trigonometry

www.akankshaclasses.com
आ
CLASS X Mathematics ~4 marks/year Ch 9 of 14
Some Applications of Trigonometry

Class 10 · Mathematics · NCERT chapter notes · Akanksha Classes

Snapshot
  • Trigonometry lets us find heights and distances we can never measure directly — towers, hills, balloons, the width of a river.
  • The line of sight joins the observer's eye to the object; the angle it makes with the horizontal is the angle of elevation (looking up) or angle of depression (looking down).
  • Every problem becomes a right triangle; pick the ratio ($\tan\theta$ usually) that connects the side you know with the side you want.
  • Key memory aid: $\tan\theta=\dfrac{\text{opposite}}{\text{adjacent}}=\dfrac{\text{height}}{\text{base}}$, and the angle of depression of B from A equals the angle of elevation of A from B (alternate angles).
  • Board weightage: ~4 marks/year — usually one 3–4 mark height-and-distance word problem with two angles ($30^\circ,45^\circ,60^\circ$).
Detailed notes

1. What this chapter is about

In Chapter 8 you learned the trigonometric ratios $\sin\theta,\ \cos\theta,\ \tan\theta$ and their values for the standard angles $30^\circ,\ 45^\circ,\ 60^\circ$. This chapter has just one section — 9.1 Heights and Distances — and it answers a very practical question: how do we measure the height of a tower, a mountain, or a tall building without climbing it?

The trick is simple. We stand at a known distance, look at the top, measure the angle our gaze makes with the ground, and let a right triangle do the rest. No new theory — only the ratios you already know, applied cleverly.

$$\sin\theta=\frac{\text{opp}}{\text{hyp}},\qquad \cos\theta=\frac{\text{adj}}{\text{hyp}},\qquad \tan\theta=\frac{\text{opp}}{\text{adj}}$$

Recall the standard values you will use constantly:

$$\sin30^\circ=\tfrac12,\ \cos30^\circ=\tfrac{\sqrt3}{2},\ \tan30^\circ=\tfrac{1}{\sqrt3};\quad \tan45^\circ=1;\quad \sin60^\circ=\tfrac{\sqrt3}{2},\ \tan60^\circ=\sqrt3$$

2. Line of sight

Imagine a student looking at the top of a minar. The straight line drawn from the eye of the observer to the point being viewed on the object is called the line of sight. In the NCERT figure, this is the line $AC$ from the eye $A$ to the top $C$ of the minar.

Everything in this chapter is built on the right triangle formed by: the line of sight (the hypotenuse), the horizontal ground (the base), and the vertical object (the height).

3. Angle of elevation

When the object is above the horizontal level — we raise our head to look at it — the angle formed by the line of sight with the horizontal is called the angle of elevation.

Formally: the angle of elevation of a point viewed is the angle formed by the line of sight with the horizontal when the point being viewed is above the horizontal level. In the minar figure, $\angle BAC$ is the angle of elevation of the top $C$ from the eye $A$.

$$\text{Look up}\;\Rightarrow\;\text{angle of \textbf{elevation}}$$

4. Angle of depression

Now picture a girl on a balcony looking down at a flower pot on the ground. Here the line of sight is below the horizontal level. The angle formed by the line of sight with the horizontal is called the angle of depression.

Formally: the angle of depression of a point viewed is the angle formed by the line of sight with the horizontal when the point is below the horizontal level — the case when we lower our head to look.

$$\text{Look down}\;\Rightarrow\;\text{angle of \textbf{depression}}$$

The most important fact in this chapter: the horizontal line from the observer's eye is parallel to the ground. So the line of sight acts as a transversal, and the angle of depression (measured from the eye) equals the angle of elevation (measured from the object) by alternate angles. This lets you move the marked angle into the triangle where it is useful.

5. The solving method — five steps

Almost every board problem yields to the same routine:

  • Step 1 — Draw the figure. Show the vertical object, the horizontal ground, the observer, and mark every angle and known length. Right-angle the corner where vertical meets horizontal.
  • Step 2 — Name the right triangle and identify which side is opposite the known angle, which is adjacent, which is the hypotenuse.
  • Step 3 — Choose the ratio that links the side you know with the side you want. Height & base together $\Rightarrow$ use $\tan\theta$ (or $\cot\theta$); a slant length (ladder, rope, string) $\Rightarrow$ use $\sin\theta$ or $\cos\theta$.
  • Step 4 — Substitute the standard value and solve for the unknown.
  • Step 5 — Add back any extra height (observer's height, building height) and state the answer with units. Rationalise surds; use $\sqrt3=1.732$ if asked for a decimal.
$$\tan\theta=\frac{\text{height of object}}{\text{horizontal distance}}$$

Two-triangle problems: when two angles are given (e.g. shadow lengths, two buildings, a flagstaff on a building), set up two equations from two right triangles that share a common side (usually the height), then eliminate to solve.

6. Worked Examples (NCERT) — fully solved

NCERT Example 1 — tower, angle $60^\circ$ from 15 m

A tower stands vertically on the ground. From a point 15 m away from its foot, the angle of elevation of the top is $60^\circ$. Find the height.

Let $AB$ be the tower, $C$ the point, with the right angle at $B$. Here height $AB$ is opposite and base $BC=15$ is adjacent, so use $\tan$:

$\tan60^\circ=\dfrac{AB}{BC}\Rightarrow \sqrt3=\dfrac{AB}{15}\Rightarrow AB=15\sqrt3$ m.

Height of the tower $=15\sqrt3$ m $\approx 25.98$ m.

NCERT Example 2 — ladder on a 5 m pole

An electrician must reach a point 1.3 m below the top of a 5 m pole, using a ladder inclined at $60^\circ$. Find the ladder's length and how far from the foot to place it. (Take $\sqrt3=1.73$.)

The point $B$ to reach is at height $BD=AD-AB=(5-1.3)=3.7$ m. The ladder $BC$ is the hypotenuse of right triangle $BDC$ (right angle at $D$).

Length: $\sin60^\circ=\dfrac{BD}{BC}\Rightarrow \dfrac{3.7}{BC}=\dfrac{\sqrt3}{2}\Rightarrow BC=\dfrac{3.7\times2}{\sqrt3}=4.28$ m (approx).

Distance of foot: $\cot60^\circ=\dfrac{DC}{BD}\Rightarrow DC=\dfrac{3.7}{\sqrt3}=2.14$ m (approx). So place the foot $2.14$ m from the pole.

NCERT Example 3 — chimney, observer 1.5 m tall

An observer 1.5 m tall is 28.5 m from a chimney. The angle of elevation of the top from her eyes is $45^\circ$. Find the height of the chimney.

Let $AB$ be the chimney, $CD$ the observer ($1.5$ m), and $DE=CB=28.5$ m the horizontal distance from the eye. We need $AB=AE+BE=AE+1.5$.

$\tan45^\circ=\dfrac{AE}{DE}\Rightarrow 1=\dfrac{AE}{28.5}\Rightarrow AE=28.5$ m.

Height of chimney $AB=(28.5+1.5)=30$ m.

NCERT Example 4 — flagstaff on a 10 m building

From point $P$ the angle of elevation of the top of a 10 m building is $30^\circ$. A flag at the top of the building makes the angle of elevation of the flag-tip $45^\circ$ from $P$. Find the flagstaff length and the distance of the building from $P$. (Take $\sqrt3=1.732$.)

Let $AB=10$ (building), $BD$ the flagstaff, $P$ the point. Two right triangles $PAB$ and $PAD$ share base $PA$.

In $\triangle PAB$: $\tan30^\circ=\dfrac{AB}{AP}\Rightarrow \dfrac{1}{\sqrt3}=\dfrac{10}{AP}\Rightarrow AP=10\sqrt3=17.32$ m. That is the distance of the building.

Let $DB=x$, so $AD=10+x$. In $\triangle PAD$: $\tan45^\circ=\dfrac{AD}{AP}\Rightarrow 1=\dfrac{10+x}{10\sqrt3}\Rightarrow 10+x=10\sqrt3$, hence $x=10(\sqrt3-1)=7.32$ m. Flagstaff $=7.32$ m.

NCERT Example 5 — shadow 40 m longer ($30^\circ$ vs $60^\circ$)

A tower's shadow on level ground is 40 m longer when the Sun's altitude is $30^\circ$ than when it is $60^\circ$. Find the height.

Let height $AB=h$ and shorter shadow $BC=x$ (at $60^\circ$); longer shadow $BD=x+40$ (at $30^\circ$).

In $\triangle ABC$: $\tan60^\circ=\dfrac{h}{x}\Rightarrow \sqrt3=\dfrac{h}{x}\Rightarrow h=x\sqrt3$ …(1)

In $\triangle ABD$: $\tan30^\circ=\dfrac{h}{x+40}\Rightarrow \dfrac{1}{\sqrt3}=\dfrac{h}{x+40}$ …(2)

Put (1) into (2): $(x\sqrt3)\sqrt3=x+40\Rightarrow 3x=x+40\Rightarrow x=20$. So $h=20\sqrt3$ m.

NCERT Example 6 — two buildings, depressions $30^\circ$ & $45^\circ$

From the top of a multi-storeyed building the angles of depression of the top and bottom of an 8 m building are $30^\circ$ and $45^\circ$. Find the height of the multi-storeyed building and the distance between them.

Let $PC$ be the tall building, $AB=8$ m the short one. Draw horizontal $PQ$ through $P$; $BD$ is horizontal at the top of the short building. By alternate angles $\angle PBD=30^\circ,\ \angle PAC=45^\circ$.

In $\triangle PBD$: $\dfrac{PD}{BD}=\tan30^\circ=\dfrac{1}{\sqrt3}\Rightarrow BD=PD\sqrt3$.

In $\triangle PAC$: $\dfrac{PC}{AC}=\tan45^\circ=1\Rightarrow PC=AC$. Since $AC=BD$ and $DC=AB=8$, and $PC=PD+DC$, we get $PD+8=PD\sqrt3$.

So $PD=\dfrac{8}{\sqrt3-1}=\dfrac{8(\sqrt3+1)}{(\sqrt3-1)(\sqrt3+1)}=4(\sqrt3+1)$. Height $PC=PD+8=\{4(\sqrt3+1)+8\}=4(3+\sqrt3)$ m, and the distance between the buildings $AC=BD=4(3+\sqrt3)$ m.

NCERT Example 7 — width of a river from a bridge

From a point $P$ on a bridge, the angles of depression of the banks on opposite sides are $30^\circ$ and $45^\circ$. The bridge is 3 m above the banks; find the river's width.

Let $A,B$ be the two banks, $D$ the foot of the perpendicular from $P$, with $DP=3$ m. Width $AB=AD+DB$. By alternate angles $\angle A=30^\circ,\ \angle B=45^\circ$.

In $\triangle APD$: $\tan30^\circ=\dfrac{PD}{AD}\Rightarrow \dfrac{1}{\sqrt3}=\dfrac{3}{AD}\Rightarrow AD=3\sqrt3$ m.

In $\triangle PBD$: $\angle B=45^\circ\Rightarrow BD=PD=3$ m.

$AB=BD+AD=3+3\sqrt3=3(1+\sqrt3)$ m. Width of river $=3(\sqrt3+1)$ m.

7. NCERT Exercise 9.1 — fully solved (Q1–Q8)

Q1. Circus rope. A 20 m rope tied from the top of a vertical pole to the ground makes $30^\circ$ with the ground. Find the pole's height. The rope is the hypotenuse: $\sin30^\circ=\dfrac{\text{height}}{20}\Rightarrow \dfrac12=\dfrac{h}{20}\Rightarrow h=10$ m.

Q2. Broken tree. The broken part makes $30^\circ$ with the ground; its foot is 8 m from the base. Let standing part $=AB$, broken part (hypotenuse) $=BC$. $\tan30^\circ=\dfrac{AB}{8}\Rightarrow AB=\dfrac{8}{\sqrt3}$. Also $\cos30^\circ=\dfrac{8}{BC}\Rightarrow BC=\dfrac{8}{\cos30^\circ}=\dfrac{16}{\sqrt3}$. Total height $=AB+BC=\dfrac{8}{\sqrt3}+\dfrac{16}{\sqrt3}=\dfrac{24}{\sqrt3}=8\sqrt3$ m.

Q3. Two slides. Younger children: top at $1.5$ m, slide at $30^\circ$. $\sin30^\circ=\dfrac{1.5}{L}\Rightarrow \dfrac12=\dfrac{1.5}{L}\Rightarrow L=3$ m. Older children: top at $3$ m, slide at $60^\circ$. $\sin60^\circ=\dfrac{3}{L}\Rightarrow \dfrac{\sqrt3}{2}=\dfrac{3}{L}\Rightarrow L=\dfrac{6}{\sqrt3}=2\sqrt3$ m.

Q4. Tower from 30 m, angle $30^\circ$. $\tan30^\circ=\dfrac{h}{30}\Rightarrow \dfrac{1}{\sqrt3}=\dfrac{h}{30}\Rightarrow h=\dfrac{30}{\sqrt3}=10\sqrt3$ m.

Q5. Kite string. Kite at height 60 m, string at $60^\circ$ (no slack). $\sin60^\circ=\dfrac{60}{L}\Rightarrow \dfrac{\sqrt3}{2}=\dfrac{60}{L}\Rightarrow L=\dfrac{120}{\sqrt3}=40\sqrt3$ m.

Q6. Boy and 30 m building. Boy's eyes at $1.5$ m, so effective height $=30-1.5=28.5$ m. Let distances be $d_1$ (at $30^\circ$) and $d_2$ (at $60^\circ$). $\tan30^\circ=\dfrac{28.5}{d_1}\Rightarrow d_1=28.5\sqrt3$; $\tan60^\circ=\dfrac{28.5}{d_2}\Rightarrow d_2=\dfrac{28.5}{\sqrt3}=9.5\sqrt3$. Distance walked $=d_1-d_2=28.5\sqrt3-9.5\sqrt3=19\sqrt3$ m.

Q7. Transmission tower on a 20 m building. Let building $=20$ m, tower $=h$, base distance $=x$. Bottom (building top) at $45^\circ$: $\tan45^\circ=\dfrac{20}{x}\Rightarrow x=20$. Top of tower at $60^\circ$: $\tan60^\circ=\dfrac{20+h}{x}\Rightarrow \sqrt3=\dfrac{20+h}{20}\Rightarrow 20+h=20\sqrt3\Rightarrow h=20(\sqrt3-1)$ m $\approx 14.64$ m.

Q8. Statue on a pedestal. Statue $1.6$ m on pedestal of height $h$; from a point, top of statue at $60^\circ$, top of pedestal at $45^\circ$. Let base distance $=x$. Pedestal: $\tan45^\circ=\dfrac{h}{x}\Rightarrow x=h$. Statue top: $\tan60^\circ=\dfrac{h+1.6}{x}\Rightarrow \sqrt3=\dfrac{h+1.6}{h}\Rightarrow \sqrt3\,h=h+1.6\Rightarrow h(\sqrt3-1)=1.6\Rightarrow h=\dfrac{1.6}{\sqrt3-1}=\dfrac{1.6(\sqrt3+1)}{2}=0.8(\sqrt3+1)$ m $\approx 2.19$ m.

8. NCERT Exercise 9.1 — fully solved (Q9–Q15)

Q9. Building and tower. Tower is 50 m. From foot of tower, top of building is $30^\circ$; from foot of building, top of tower is $60^\circ$. Let building $=h$, distance $=d$. From foot of building: $\tan60^\circ=\dfrac{50}{d}\Rightarrow d=\dfrac{50}{\sqrt3}$. From foot of tower: $\tan30^\circ=\dfrac{h}{d}\Rightarrow \dfrac{1}{\sqrt3}=\dfrac{h}{50/\sqrt3}\Rightarrow h=\dfrac{50}{3}=16\tfrac{2}{3}$ m.

Q10. Two equal poles, road 80 m wide. A point between them sees the tops at $60^\circ$ and $30^\circ$. Let pole height $=h$, distance from point to nearer pole $=x$, so to farther pole $=80-x$. $\tan60^\circ=\dfrac{h}{x}\Rightarrow h=x\sqrt3$; $\tan30^\circ=\dfrac{h}{80-x}\Rightarrow h=\dfrac{80-x}{\sqrt3}$. Equate: $x\sqrt3=\dfrac{80-x}{\sqrt3}\Rightarrow 3x=80-x\Rightarrow x=20$. So $h=20\sqrt3$ m. The point is $20$ m from one pole and $60$ m from the other.

Q11. TV tower on a canal bank. From a point opposite, elevation $60^\circ$; from a point 20 m further back, elevation $30^\circ$. Let tower height $=h$, canal width (nearer point to foot) $=x$. $\tan60^\circ=\dfrac{h}{x}\Rightarrow h=x\sqrt3$; $\tan30^\circ=\dfrac{h}{x+20}\Rightarrow h=\dfrac{x+20}{\sqrt3}$. Equate: $x\sqrt3=\dfrac{x+20}{\sqrt3}\Rightarrow 3x=x+20\Rightarrow x=10$. Width of canal $=10$ m, height of tower $=10\sqrt3$ m.

Q12. Cable tower from a 7 m building. From the top of a 7 m building, top of cable tower at elevation $60^\circ$, foot at depression $45^\circ$. The depression gives horizontal distance: $\tan45^\circ=\dfrac{7}{d}\Rightarrow d=7$ m. The elevation gives the part of the tower above building level: $\tan60^\circ=\dfrac{a}{d}\Rightarrow a=7\sqrt3$. Total tower height $=7+7\sqrt3=7(1+\sqrt3)$ m.

Q13. Two ships from a 75 m lighthouse. Depressions $30^\circ$ and $45^\circ$, ships on the same side. Nearer ship ($45^\circ$): $\tan45^\circ=\dfrac{75}{d_1}\Rightarrow d_1=75$ m. Farther ship ($30^\circ$): $\tan30^\circ=\dfrac{75}{d_2}\Rightarrow d_2=75\sqrt3$ m. Distance between ships $=d_2-d_1=75\sqrt3-75=75(\sqrt3-1)$ m.

Q14. Balloon moving. Girl 1.2 m tall; balloon at height $88.2$ m, so above eye level $=88.2-1.2=87$ m. Elevation drops from $60^\circ$ to $30^\circ$. First horizontal distance: $\tan60^\circ=\dfrac{87}{x_1}\Rightarrow x_1=\dfrac{87}{\sqrt3}=29\sqrt3$. Later: $\tan30^\circ=\dfrac{87}{x_2}\Rightarrow x_2=87\sqrt3$. Distance travelled $=x_2-x_1=87\sqrt3-29\sqrt3=58\sqrt3$ m.

Q15. Car approaching the tower. From the top of the tower, depression of the car is $30^\circ$; 6 seconds later it is $60^\circ$. Let tower height $=h$. At $30^\circ$ the horizontal distance is $h\cot30^\circ=h\sqrt3$; at $60^\circ$ it is $h\cot60^\circ=\dfrac{h}{\sqrt3}$. Distance covered in 6 s $=h\sqrt3-\dfrac{h}{\sqrt3}=\dfrac{2h}{\sqrt3}$. Remaining distance to the foot $=\dfrac{h}{\sqrt3}$. Time $\propto$ distance at constant speed: remaining time $=\dfrac{h/\sqrt3}{2h/\sqrt3}\times6=\dfrac{1}{2}\times6=3$ seconds.

9. Common mistakes to avoid

  • Confusing elevation (look up) with depression (look down) — read whether the observer is above or below the object.
  • Forgetting that the angle of depression equals the angle of elevation by alternate angles; students wrongly place the depression angle inside the lower triangle.
  • Using $\sin$/$\cos$ when only height and base are involved — those need $\tan$ (or $\cot$). Use $\sin$/$\cos$ only when a slant length (rope, ladder, string) appears.
  • Forgetting to add the observer's height or the building's height back at the end (Examples 3, 4).
  • Mixing up which shadow/distance is longer — the smaller angle gives the longer horizontal distance.
  • Leaving answers un-rationalised, e.g. writing $\dfrac{30}{\sqrt3}$ instead of $10\sqrt3$.
  • Not drawing the right angle at the foot — every triangle here is right-angled where the vertical meets the ground.

10. Quick revision checklist

  • Line of sight = eye to object; elevation = look up, depression = look down.
  • Angle of depression of B from A $=$ angle of elevation of A from B (alternate angles).
  • Height & base $\Rightarrow$ $\tan\theta=\dfrac{\text{height}}{\text{base}}$. Slant length $\Rightarrow$ $\sin\theta$ or $\cos\theta$.
  • Two angles given $\Rightarrow$ two right triangles sharing the height; form two equations and eliminate.
  • Standard values: $\tan30^\circ=\tfrac{1}{\sqrt3},\ \tan45^\circ=1,\ \tan60^\circ=\sqrt3$.
  • Always draw the figure, mark the right angle, add back extra heights, rationalise, state units.
Practice MCQs
1. The angle of elevation of the top of a tower from a point 100 m away (on level ground) is $45^\circ$. The height of the tower is:
  1. $50$ m
  2. $100$ m
  3. $100\sqrt3$ m
  4. $\dfrac{100}{\sqrt3}$ m
Answer: (B) $\tan45^\circ=\dfrac{h}{100}=1\Rightarrow h=100$ m.
2. The line drawn from the eye of an observer to the point viewed on the object is called the:
  1. angle of elevation
  2. horizontal level
  3. line of sight
  4. angle of depression
Answer: (C) line of sight.
3. If the height of a tower equals the length of its shadow, the Sun's angle of elevation is:
  1. $30^\circ$
  2. $45^\circ$
  3. $60^\circ$
  4. $90^\circ$
Answer: (B) $\tan\theta=\dfrac{h}{h}=1\Rightarrow\theta=45^\circ$.
4. A 20 m ladder leans against a wall making $60^\circ$ with the ground. The height reached on the wall is:
  1. $10$ m
  2. $10\sqrt3$ m
  3. $20\sqrt3$ m
  4. $\dfrac{20}{\sqrt3}$ m
Answer: (B) $\sin60^\circ=\dfrac{h}{20}=\dfrac{\sqrt3}{2}\Rightarrow h=10\sqrt3$ m.
5. The angle of depression of an object from the top of a tower is the angle the line of sight makes with the:
  1. vertical tower
  2. horizontal level
  3. object's base
  4. ground at the foot
Answer: (B) the horizontal level through the observer's eye.
6. From 30 m away, the elevation of the top of a tower is $30^\circ$. The height of the tower is:
  1. $30\sqrt3$ m
  2. $10\sqrt3$ m
  3. $15$ m
  4. $\dfrac{30}{\sqrt3}$ m only
Answer: (B) $\tan30^\circ=\dfrac{h}{30}=\dfrac{1}{\sqrt3}\Rightarrow h=\dfrac{30}{\sqrt3}=10\sqrt3$ m.
7. As the angle of elevation of the Sun decreases, the length of a tower's shadow:
  1. increases
  2. decreases
  3. stays the same
  4. becomes zero
Answer: (A) smaller angle $\Rightarrow$ larger $\cot\theta\Rightarrow$ longer shadow.
8. A kite is at height 75 m; the string makes $60^\circ$ with the ground (no slack). The string length is:
  1. $50\sqrt3$ m
  2. $75\sqrt3$ m
  3. $\dfrac{150}{\sqrt3}$ m
  4. both A and C
Answer: (D) $\sin60^\circ=\dfrac{75}{L}\Rightarrow L=\dfrac{150}{\sqrt3}=50\sqrt3$ m — A and C are equal.
9. The angle of elevation of the top of a tower from a point increases as the point:
  1. moves away from the foot
  2. moves towards the foot
  3. stays fixed
  4. moves up
Answer: (B) closer to the foot $\Rightarrow$ larger angle of elevation.
10. The tops of two poles of equal height, $80$ m apart on opposite sides of a road, subtend $60^\circ$ and $30^\circ$ at a point between them. The height of the poles is:
  1. $20$ m
  2. $20\sqrt3$ m
  3. $40$ m
  4. $40\sqrt3$ m
Answer: (B) solving gives the nearer distance $20$ m, so $h=20\sqrt3$ m (Ex. 9.1 Q10).
Assertion–Reason
A: The angle of depression of a point B seen from A equals the angle of elevation of A seen from B.   R: A horizontal line through the eye is parallel to the ground, so alternate angles are equal.
Answer: Both A and R are true, and R is the correct explanation of A — the parallel horizontals make the line of sight a transversal with equal alternate angles.
A: To find a tower's height from its base distance and the angle of elevation, we use $\sin\theta$.   R: $\sin\theta=\dfrac{\text{opposite}}{\text{hypotenuse}}$.
Answer: A is false (height and base together need $\tan\theta$, not $\sin\theta$), R is true. So A is false, R is true.
Previous-year questions
Q1. The shadow of a tower is $40$ m longer when the Sun's altitude is $30^\circ$ than when it is $60^\circ$. Find the height of the tower. (CBSE, 3 marks)
Answer: $h=x\sqrt3$ and $h=\dfrac{x+40}{\sqrt3}\Rightarrow 3x=x+40\Rightarrow x=20$, so $h=20\sqrt3$ m (NCERT Example 5).
Q2. From the top of a 7 m building, the angle of elevation of the top of a cable tower is $60^\circ$ and the angle of depression of its foot is $45^\circ$. Find the height of the tower. (CBSE, 4 marks)
Answer: Depression $\Rightarrow$ distance $=7$ m; elevation $\Rightarrow$ upper part $=7\sqrt3$. Height $=7+7\sqrt3=7(1+\sqrt3)$ m (Ex. 9.1 Q12).
Q3. A man on the top of a tower observes a car at depression $30^\circ$ approaching at uniform speed; 6 s later the depression is $60^\circ$. Find the time taken by the car to reach the foot from this point. (CBSE, 4 marks)
Answer: Distance covered in 6 s $=\dfrac{2h}{\sqrt3}$, remaining $=\dfrac{h}{\sqrt3}$, so remaining time $=\dfrac12\times6=3$ seconds (Ex. 9.1 Q15).
Q4. The angles of depression of the top and bottom of an 8 m tall building from the top of a multi-storeyed building are $30^\circ$ and $45^\circ$. Find the height of the multi-storeyed building. (CBSE, 4 marks)
Answer: $PD=\dfrac{8}{\sqrt3-1}=4(\sqrt3+1)$, height $=PD+8=4(3+\sqrt3)$ m (NCERT Example 6).
Want personal coaching in Dwarka?
Book a free demo class
More Class 10 Mathematics chapters
Real Numbers · Polynomials · Pair of Linear Equations in Two Variables · Quadratic Equations · Arithmetic Progressions · Triangles · Coordinate Geometry · Introduction to Trigonometry