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
+977-10-4382404, 01-4362082
vedantapublication@gmail.com


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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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