Exercise 16.1

Other solutions from this unit:

General Section


1. Find the unknown lengths of the sides of the following figures. [Refer to your book for the figures]


a) 
Solution:

Here,
$tan45 ^o = \dfrac{x}{20}$

$or, 1 = \dfrac{x}{20}$

$\therefore x = 20m$

And,
$sin 45 ^o = \dfrac{x}{y}$

$or, \dfrac{1}{\sqrt{2}} = \dfrac{20}{y}$

$\therefore y = 20 \sqrt{2} m$


b)
Solution:

Here,
$tan 60 ^o = \dfrac{40 \sqrt{3}}{x}$

$or, \sqrt{3} = \dfrac{40 \sqrt{3}}{x}$

$\therefore x = 40 m$

And,

$sin 60 ^o = \dfrac{40 \sqrt{3}}{y}$

$or, \dfrac{\sqrt{3}}{2} = \dfrac{40 \sqrt{3}}{y}$

$or, y = 40 * 2$

$\therefore y = 80m$

c)
Solution:

Here,
$sin 30 ^o = \dfrac{PQ}{PR} = \dfrac{x}{150}$

$or, \dfrac{1}{2} = \dfrac{x}{150}$

$or, x = \dfrac{150}{2}$

$\therefore x = 75m$

And,
$cos 30 ^o = \dfrac{QR}{PR} = \dfrac{y}{150}$

$or, \dfrac{\sqrt{3}}{2} = \dfrac{y}{150}$

$or, y = \dfrac{ 150 \sqrt{3}}{2}$

$\therefore y = 75 \sqrt{3} m$


e)
Solution:

Here,
In rectangle ABDE,
$AB = ED = 1.5m$
$AE = BD = xm$

And,
$CD = CE + ED$

$or, CE = CD - ED$

$or, CE = 121.5 - 1.5$

$\therefore CE = 120m$

Now,
$tan 60 ^o = \dfrac{CE}{AE} = \dfrac{120}{x}$

$or, \sqrt{3} = \dfrac{120}{x}$

$or, x = \dfrac{ 40 * \sqrt{3} * \sqrt{3}}{\sqrt{3}}$

$\therefore x = 40 \sqrt{3}m$

g)
Solution:

Here,
In rectangle PQST,
$PQ = TS = xm$
$QS = PT = 80m$

We know,
$RS = RT + TS$

$or, RT = RS - TS$

$or, RT = (200 - x)m$

In triangle PRT,
$tan 45 ^o = \dfrac{RT}{PT} = \dfrac{200 - x}{80}$

$or, 1 = \dfrac{ 200 - x}{80}$

$or, 80 = 200 - x$

$or, x = 200 - 80$

$\therefore x = 120m$


2 a)


2 c)


Creative Section

3 a) The angle of elevation of the top of a tree observed from a point 60m away from its foot is 45 degree. Find the height of the tree.

Solution:

[You must draw a figure. We haven't drawn it here.]

Letthe height of the tree be 'h'.

Here,
$tan 45^o = \dfrac{height}{distance}$

$or, 1 = \dfrac{h}{60}$

$\therefore h = 60m$

Hence, the required height of the tree is 60 m.

3






4 a) A circular pond has a pole standing vertically at its centre. The top of the pole is 30m above the water surface and the angle of elevation of it from a point on the circumference is 60°. Find the length of radius of the pond.

Solution:

[Figure drawing is compulsory.]

Let the length of radius of the pond be 'r'.

Here,
height of the pole (h) = 30m
angle of elevation ($\theta$) = 60°$

According to the question,
$tan \theta = \dfrac{height}{radius}$

$or, tan 60° = \dfrac{30}{r}$

$or, r = \dfrac{30}{tan60°}$

$or, r = \dfrac{10 × √3 × √3}{√3}$

$\therefore r = 10√3m$

Hence, the required radius of the pond is 10√3 m.




5 a) An observer finds the angle of elevation of the top of a pole to be 60°. If the height of the pole and the observer are 25.3 m and 1.3 m respectively, find the distance between the observer and the pole.

Solution:




Here,
Height of the pole as observed by the observer (height) (AE) = height of pole - height of observer

$= BE - AB$

$= BE - CD$ [AB = CD, opposite sides of a rectangle]

$= 25.3 - 1.3$

$= 24 m$

Given,
angle of elevation ($\theta$) = 60°

According to the question and figure,

$tan \theta = \dfrac{height}{distance}$

$or, tan 60° = \dfrac{24}{distance}$

$or, distance = \dfrac{24}{tan60°}$

$or, distance = \dfrac{8×√3×√3}{√3}$

$\therefore distance = 8√3$

Hence, the required distance between the observer and the tower is 8√3m.


6 a) On a windy day, a girl of height 1.1 m was flying her kite. When the length of the string of the kite was 33 metres, it makes an angle of 30° with horizon. At what height was the kite above the ground?

Solution:

You must draw a figure.

Here,
angle of elevation ($\theta$) = 30°
length of string of kite (l) = 33 metres
apparent height of kite (h) = m

From given and figure,

$cos \theta = \dfrac{h}{l}$

$or, sin 30° = \dfrac{x}{33}$

$or, \dfrac{1}{2} = \dfrac{x}{33}$

$or, \dfrac{33}{2} = x$

$\therefore x = 16.5 m$

Now,
Actual height of the kite =( x + 1.1) m

$= (16.5+1.1)m$

$= 17.6 m$

Hence, the required height of the kite above the ground is 17.6 m.


[We added the height of the girl to the height of the kite because we wanted to find the actual height of the kite above the ground not above the girl.]


7 a) From the top of a building 20m high, a 1.7 m tall man observes the elevation of the top of a tower and finds it 45°. If the distance between the building and the tower is 50 m, find the height of the tower.

Solution:

Here,
distance between tower and building (b) = 50 m
angle of elevation ($\theta$) = 45°
height of building (h1) = 20m
height of man (h2) = 1.7m

Let the observed height of the tower be 'h3'.

Now,
$tan \theta = \dfrac{h3}{b}$

$or, tan 45° = \dfrac{h3}{50}$

$or, 1 = \dfrac{h3}{50}$

$\therefore h3 = 50 m$

And,

The required height of the tower = $h1 + h2 + h3$

$= 20 + 1.7 + 50$

$= 71.7 m$

Hence, the required height of the tower is 71.7 m.


8 a) A man is 2 m tall and the length of his shadow in the sun is 2√3 m. Find the altitude of the sun.

Solution:
Here, altitude of the sun refers to the angle of elevation of the sun.

Given,
height of man (p) = 2m
length of shadow (b) = 2√3m

Let the altitude of the sun be $\theta$

$or, tan \theta = \dfrac{p}{b}$

$or, tan \theta = \dfrac{2}{2√3}$

$or, tan \theta = \dfrac{1}{√3}$

$or, \theta = tan^{-1} \left ( \dfrac{1}{√3} \right )$

$\therefore \theta = 30°$

Hence, the required altitude of the sun is 30°.


9 a) The upper part of a straight tree broken by the wind makes an angle of 45° with the plane surface at a point 9 m from the foot of the tree. Find the height of the tree before it was broken.

Solution:

Let the total height of the tree be H.

Angle of elevation ($\theta$) = 45°
Distance between foot of tree and broken part (b) = 9m

Let the height of the part of tree attached to the base be p.

Let the length of the broken part of the tree be h.

From given,
$tan \theta = \dfrac{p}{b}$

$or, tan 45° = \dfrac{p}{9}$

$\therefore p = 9m$

And,
$sin \theta = \dfrac{p}{h}$

$or, sin 45° = \dfrac{9}{h}$

$or, \dfrac{1}{√2} = \dfrac{9}{h}$

$\therefore h = 9√2m$

Again,
H = p + h

$= 9 + 9√2$

$= 9 + 12.72$

$= 21.72 m$

Hence, the required height of the tree before it was broken was 21.72 m.

















About vedanta EXCEL in MATHEMATICS Book 10

Author: Hukum Pd. Dahal
Editor: Tara Bahadur Magar

Vanasthali, Kathmandu, Nepal
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Exercise 16.1 | Height and Distance | vedanta Excel in Mathematics | Class 10 is a collection of the solutions related to exercises of height and distance from Trigonometry Chapter for Nepal's Secondary Education Examination (SEE) appearing students.

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