Some Applications of Trigonometry — CBSE Class 10 Maths Textbook Solutions

Let $AB$AB be the pole (height $h$h) and $C$C the point where the rope meets the ground, so the rope $AC = 20$AC = 20 m is the hypotenuse of right triangle $ABC$ABC, right-angled at $B$B, with $\angle ACB = 30^\circ$ACB = 30^.

Since the height $AB$AB is opposite $\angle C$C and the rope $AC$AC is the hypotenuse, use $\sin$:
$\sin30^\circ=\frac{AB}{AC}=\frac{h}{20}$30^=AB/AC=h/20
$\frac12=\frac{h}{20} \implies h=10$12=h/20 h=10

The height of the pole is $10$10 m.

Let the tree originally stand at $A$A (foot), and break at point $B$B, height $h$h above the ground. The broken part falls to touch the ground at $C$C, with $AC = 8$AC = 8 m, and $\angle ACB = 30^\circ$ACB = 30^. Triangle $ABC$ABC is right-angled at $A$A.

Height of the stump $AB$AB: opposite to $\angle C$C, adjacent side $AC=8$AC=8 known, so use $\tan$:
$\tan30^\circ=\frac{AB}{AC}=\frac{h}{8}\implies h=8\times\frac{1}{\sqrt3}=\frac{8}{\sqrt3}=\frac{8\sqrt3}{3}\text{ m}$30^=AB/AC=h/8 h=8×1/3=8/3=83/3 m

Length of the broken part $BC$BC (hypotenuse):
$\cos30^\circ=\frac{AC}{BC}=\frac{8}{BC}\implies BC=\frac{8}{\cos30^\circ}=\frac{8}{\sqrt3/2}=\frac{16}{\sqrt3}=\frac{16\sqrt3}{3}\text{ m}$30^=AC/BC=8/BC BC=8/30^=8/3/2=16/3=163/3 m

Total height of the tree = stump + broken part:
$h+BC=\frac{8}{\sqrt3}+\frac{16}{\sqrt3}=\frac{24}{\sqrt3}=\frac{24\sqrt3}{3}=8\sqrt3\text{ m}$h+BC=8/3+16/3=24/3=243/3=83 m

The height of the tree is $8\sqrt3$83 m.

In each case, the slide is the hypotenuse of a right triangle whose opposite side (to the angle of inclination at the ground) is the given height.

Younger children's slide (height $1.5$1.5 m, angle $30^\circ$30^):
$\sin30^\circ=\frac{1.5}{\text{length}}\implies \text{length}=\frac{1.5}{\sin30^\circ}=\frac{1.5}{1/2}=3\text{ m}$30^=1.5/length length=1.5/30^=1.5/1/2=3 m

Elder children's slide (height $3$3 m, angle $60^\circ$60^):
$\sin60^\circ=\frac{3}{\text{length}}\implies \text{length}=\frac{3}{\sin60^\circ}=\frac{3}{\sqrt3/2}=\frac{6}{\sqrt3}=2\sqrt3\text{ m}$60^=3/length length=3/60^=3/3/2=6/3=23 m

The slide lengths are $3$3 m (for younger children) and $2\sqrt3$23 m (for elder children).

Let $AB$AB be the tower (height $h$h) and $C$C the point on the ground, $BC = 30$BC = 30 m, with $\angle ACB = 30^\circ$ACB = 30^.

Since the height $AB$AB is opposite the angle and $BC$BC is adjacent, use $\tan$:
$\tan30^\circ=\frac{AB}{BC}=\frac{h}{30}$30^=AB/BC=h/30
$\frac{1}{\sqrt3}=\frac{h}{30}\implies h=\frac{30}{\sqrt3}=10\sqrt3\text{ m}$1/3=h/30 h=30/3=103 m

The height of the tower is $10\sqrt3$103 m.

The height of the kite ($60$60 m) is the side opposite the $60^\circ$60^ angle, and the string is the hypotenuse.

$\sin60^\circ=\frac{60}{\text{string length}}$60^=60/string length
$\frac{\sqrt3}{2}=\frac{60}{L}\implies L=\frac{60\times2}{\sqrt3}=\frac{120}{\sqrt3}=40\sqrt3\text{ m}$3/2=60/L L=60×2/3=120/3=403 m

The length of the string is $40\sqrt3$403 m.

The relevant vertical height (above the boy's eye level) is $30-1.5=28.5$30-1.5=28.5 m, constant throughout.

Let $d_1$d_1 = distance from the far point (angle $30^\circ$30^), $d_2$d_2 = distance from the near point (angle $60^\circ$60^).
$\tan30^\circ=\frac{28.5}{d_1}\implies d_1=28.5\sqrt3$30^=28.5/d_1 d_1=28.53
$\tan60^\circ=\frac{28.5}{d_2}\implies d_2=\frac{28.5}{\sqrt3}=\frac{28.5\sqrt3}{3}=9.5\sqrt3$60^=28.5/d_2 d_2=28.5/3=28.53/3=9.53

Distance walked $=d_1-d_2=28.5\sqrt3-9.5\sqrt3=19\sqrt3$=d_1-d_2=28.53-9.53=193 m.

The boy walked $19\sqrt3$193 m towards the building.

Let $AB=20$AB=20 m be the building and $BC=x$BC=x be the tower on top of it, observed from point $P$P on the ground, with $PA=$PA= horizontal distance.

Using the $45^\circ$45^ angle (to the foot of the tower, $B$B):
$\tan45^\circ=\frac{AB}{PA}=\frac{20}{PA}\implies 1=\frac{20}{PA}\implies PA=20\text{ m}$45^=AB/PA=20/PA 1=20/PA PA=20 m

Using the $60^\circ$60^ angle (to the top of the tower, $C$C):
$\tan60^\circ=\frac{AB+BC}{PA}=\frac{20+x}{20}$60^=AB+BC/PA=20+x/20
$\sqrt3=\frac{20+x}{20}\implies 20+x=20\sqrt3\implies x=20\sqrt3-20=20(\sqrt3-1)$3=20+x/20 20+x=203 x=203-20=20(3-1)

The height of the tower is $20(\sqrt3-1)$20(3-1) m.

Let the pedestal height be $h$h and the horizontal distance from the observation point be $d$d.

Using the $45^\circ$45^ angle (top of pedestal):
$\tan45^\circ=\frac{h}{d}\implies 1=\frac{h}{d}\implies d=h$45^=h/d 1=h/d d=h

Using the $60^\circ$60^ angle (top of statue):
$\tan60^\circ=\frac{h+1.6}{d}=\frac{h+1.6}{h}$60^=h+1.6/d=h+1.6/h
$\sqrt3=\frac{h+1.6}{h}\implies \sqrt3\,h=h+1.6\implies h(\sqrt3-1)=1.6$3=h+1.6/h 3\,h=h+1.6 h(3-1)=1.6
$h=\frac{1.6}{\sqrt3-1}=\frac{1.6(\sqrt3+1)}{(\sqrt3-1)(\sqrt3+1)}=\frac{1.6(\sqrt3+1)}{2}=0.8(\sqrt3+1)$h=1.6/3-1=1.6(3+1)/(3-1)(3+1)=1.6(3+1)/2=0.8(3+1)

The height of the pedestal is $0.8(\sqrt3+1)$0.8(3+1) m (approximately $2.19$2.19 m).

Let the building $AB=h$AB=h and the tower $CD=50$CD=50 m stand on the same level ground, distance $BC=AD=d$BC=AD=d apart (the horizontal distance is the same segment viewed from both feet).

Let $d$d be the horizontal distance between the tower and building.

From the foot of the tower, looking at the building (angle $30^\circ$30^):
$\tan30^\circ=\frac{h}{d} \implies h=\frac{d}{\sqrt3}$30^=h/d h=d/3

From the foot of the building, looking at the tower (angle $60^\circ$60^):
$\tan60^\circ=\frac{50}{d}\implies d=\frac{50}{\sqrt3}$60^=50/d d=50/3

Substituting:
$h=\frac{1}{\sqrt3}\times\frac{50}{\sqrt3}=\frac{50}{3}=16\frac23\text{ m}$h=1/3×50/3=50/3=1623 m

The height of the building is $\dfrac{50}{3}$50/3 m $=16\dfrac23$=1623 m.

Let the equal pole height be $h$h, and let the point on the road be $x$x m from the pole seen at $60^\circ$60^, so it is $(80-x)$(80-x) m from the other pole (seen at $30^\circ$30^).

From the point, angle $60^\circ$60^ to the nearer pole (distance $x$x):
$\tan60^\circ=\frac{h}{x}\implies h=x\sqrt3$60^=h/x h=x3

Angle $30^\circ$30^ to the farther pole (distance $80-x$80-x):
$\tan30^\circ=\frac{h}{80-x}\implies h=\frac{80-x}{\sqrt3}$30^=h/80-x h=80-x/3

Equating the two expressions for $h$h:
$x\sqrt3=\frac{80-x}{\sqrt3}\implies 3x=80-x\implies 4x=80\implies x=20$x3=80-x/3 3x=80-x 4x=80 x=20
$h=20\sqrt3$h=203

The height of each pole is $20\sqrt3$203 m, and the point is $20$20 m from one pole and $60$60 m from the other.

Let the tower height be $h$h, and let $d$d be the width of the canal (distance from the near bank point to the foot of the tower).

From the point directly opposite (distance $d$d), angle $60^\circ$60^:
$\tan60^\circ=\frac{h}{d}\implies h=d\sqrt3$60^=h/d h=d3

From the point $20$20 m further back (distance $d+20$d+20), angle $30^\circ$30^:
$\tan30^\circ=\frac{h}{d+20}\implies h=\frac{d+20}{\sqrt3}$30^=h/d+20 h=d+20/3

Equating:
$d\sqrt3=\frac{d+20}{\sqrt3}\implies 3d=d+20\implies 2d=20\implies d=10$d3=d+20/3 3d=d+20 2d=20 d=10
$h=10\sqrt3$h=103

The height of the tower is $10\sqrt3$103 m, and the width of the canal is $10$10 m.

Let $A$A be the top of the $7$7 m building, $B$B its foot. Let $C$C be the foot of the cable tower and $D$D its top; the horizontal distance between the building and the tower is $x$x.

Angle of depression $45^\circ$45^ to the foot $C$C: this equals the angle of elevation of $A$A from $C$C (alternate angles), so in the right triangle with horizontal leg $x$x and vertical leg $AB=7$AB=7 m:
$\tan45^\circ=\frac{AB}{x}=\frac{7}{x}\implies 1=\frac{7}{x}\implies x=7$45^=AB/x=7/x 1=7/x x=7

Angle of elevation $60^\circ$60^ to the top $D$D: let the height of $D$D above $A$A's level be $y$y.
$\tan60^\circ=\frac{y}{x}=\frac{y}{7}\implies \sqrt3=\frac{y}{7}\implies y=7\sqrt3$60^=y/x=y/7 3=y/7 y=73

Total height of the tower $CD = y+AB=7\sqrt3+7=7(\sqrt3+1)$CD = y+AB=73+7=7(3+1) m.

The height of the cable tower is $7(\sqrt3+1)$7(3+1) m.

Let $A$A be the top of the lighthouse, $B$B its foot ($AB=75$AB=75 m). Let $C$C be the nearer ship (angle of depression $45^\circ$45^) and $D$D the farther ship (angle of depression $30^\circ$30^), both on the same side.

Since the angle of depression equals the alternate angle of elevation at each ship:

Nearer ship $C$C (angle $45^\circ$45^):
$\tan45^\circ=\frac{75}{BC}\implies BC=75\text{ m}$45^=75/BC BC=75 m

Farther ship $D$D (angle $30^\circ$30^):
$\tan30^\circ=\frac{75}{BD}\implies BD=\frac{75}{\tan30^\circ}=75\sqrt3\text{ m}$30^=75/BD BD=75/30^=753 m

Distance between the ships $= BD-BC=75\sqrt3-75=75(\sqrt3-1)$= BD-BC=753-75=75(3-1) m.

The distance between the two ships is $75(\sqrt3-1)$75(3-1) m.

The height of the balloon above the girl's eye level is constant: $88.2-1.2=87$88.2-1.2=87 m.

At $60^\circ$60^, horizontal distance $d_1$d_1:
$\tan60^\circ=\frac{87}{d_1}\implies d_1=\frac{87}{\sqrt3}=29\sqrt3\text{ m}$60^=87/d_1 d_1=87/3=293 m

At $30^\circ$30^, horizontal distance $d_2$d_2:
$\tan30^\circ=\frac{87}{d_2}\implies d_2=87\sqrt3\text{ m}$30^=87/d_2 d_2=873 m

Distance travelled by the balloon $=d_2-d_1=87\sqrt3-29\sqrt3=58\sqrt3$=d_2-d_1=873-293=583 m.

The balloon travelled $58\sqrt3$583 m.

Let the tower height be $h$h. Let $d_1$d_1 be the car's distance from the foot of the tower at the $30^\circ$30^ observation, and $d_2$d_2 its distance at the $60^\circ$60^ observation (6 seconds later).

$\tan30^\circ=\frac{h}{d_1}\implies d_1=h\sqrt3$30^=h/d_1 d_1=h3
$\tan60^\circ=\frac{h}{d_2}\implies d_2=\frac{h}{\sqrt3}$60^=h/d_2 d_2=h/3

The car covers $d_1-d_2=h\sqrt3-\dfrac{h}{\sqrt3}=\dfrac{3h-h}{\sqrt3}=\dfrac{2h}{\sqrt3}$d_1-d_2=h3-h/3=3h-h/3=2h/3 in $6$6 seconds, so its (uniform) speed is
$\text{speed}=\frac{2h/\sqrt3}{6}=\frac{h}{3\sqrt3}\text{ per second}$speed=2h/3/6=h/33 per second

From the $60^\circ$60^ position, the remaining distance to the foot of the tower is $d_2=\dfrac{h}{\sqrt3}$d_2=h/3. Time taken:
$\text{time}=\frac{d_2}{\text{speed}}=\frac{h/\sqrt3}{h/(3\sqrt3)}=\frac{h}{\sqrt3}\times\frac{3\sqrt3}{h}=3$time=d_2/speed=h/3/h/(33)=h/3×33/h=3

The car takes $3$3 seconds to reach the foot of the tower from the $60^\circ$60^ position.