Natural Resource and Human Dependency

Living beings survives in water, air, forests, minerals and land. Humans cultivate wheat, maize, mustard and barley. This note provides us an information about natural resources on which human depend on.

Summary

Living beings survives in water, air, forests, minerals and land. Humans cultivate wheat, maize, mustard and barley. This note provides us an information about natural resources on which human depend on.

Things to Remember

  • Living beings survives in water, air, forests, minerals and land.
  • Humans cultivate wheat, maize, mustard and barley.
  • Harro, barro, bhojo, sarpagandha, jatamasi (spikenard), panchaunle, yachagumba are important medicinal herbs.
  • Humans use wind energy in construction and transportation.
  • Water is used to drink, clean, irrigate, bath and so on.
  • Most of the livings beings lives on land. Food and habitat of human and many animals depend on the land.

MCQs

No MCQs found.

Subjective Questions

Q1:

In a right angled triangle ABC, right angled at B, tanA = \(\frac{4}{3}\) and AB = 6cm, find the lenght AC.


Type: Short Difficulty: Easy

Show/Hide Answer
Answer: <p>Solution:</p> <p>In the right angled &Delta;ABC, AB = 6cm, tanA = \(\frac{4}{3}\)</p> <p>Now,</p> <p>or, tanA = \(\frac{BC}{AB}\)</p> <p>or, \(\frac{4}{3}\) = \(\frac{BC}{6}\)</p> <p>or, 3BC = 24</p> <p>or, BC = \(\frac{24}{3}\) = 8cm</p> <p>Also,</p> <p>AC = \(\sqrt{AB^2+BC^2}\)</p> <p>= \(\sqrt{6^2+8^2}\)</p> <p>= \(\sqrt{36+64}\)</p> <p>= \(\sqrt{100}\)</p> <p>= 10</p> <p>&there4; AC = 10cm</p> <p></p> <p></p>

Q2:

In a right angled triangle ABC, ∠O = 90°, OQ = \(\sqrt{8}\), and PO = 10cm, find the remaininig parts of the triangle.


Type: Short Difficulty: Easy

Show/Hide Answer
Answer: <p>Solution:</p> <p>Here, PO = 10cm, OQ = \(\sqrt{2}\) and &ang;O = 90&deg;.<br>By Pythagoras theorem, we have,<br>or, PQ<sup>2</sup> = PO<sup>2</sup> + OQ<sup>2</sup><br>or, PQ<sup>2 </sup>=10<sup>2</sup> + (\(\sqrt{8}\))<sup>2<br></sup>or, PQ<sup>2</sup>= 100 + 8<br>or, PQ = \(\sqrt{108}\)<br>or, PQ = \(\sqrt{36 &times; 3}\)<br>or, PQ = \(\sqrt{6^2 &times; 3}\)<br>or, PQ = 6 \(\sqrt{3}\)<br>&there4; PQ = 6 \(\sqrt{3}\) cm<br><br><br></p>

Q3:

In a right angled triangle MNO, ∠O = 90°, ∠N = 30° and n = 2\(\sqrt{3}\)cm. Solve ΔMNO.


Type: Long Difficulty: Easy

Show/Hide Answer
Answer: <p>Solution:</p> <p>Here,<br>&ang;O = 90&deg;, &ang;N = 30&deg; and n = 2\(\sqrt{3}\)cm<br>NO = ? and NM = ?</p> <p>From the right angled&Delta;MNO, we have<br>or, &ang;N + &ang;M = 90&deg;<br>or, 30&deg; + &ang;M = 90&deg;<br>or, &ang;M = 90&deg;&minus; 30&deg;<br>or, &ang;M = 60&deg;<br>Now,<br>or, tan30&deg; = \(\frac{OM}{NO}\)<br>or, \(\frac{1}{\sqrt{3}}\) = \(\frac{2\sqrt{3}}{m}\)<br>or, m = 2&times; 3<br>&there4; m = 6cm<br>Again,<br>or, sin30&deg; = \(\frac{OM}{NM}\)<br>or, \(\frac{1}{2}\) = \(\frac{2\sqrt{3}}{o}\)<br>or, o = 4\(\sqrt{3}\)<br>Hence, o = 4\(\sqrt{3}\), m = 6cm and &ang;M = 60&deg;</p>

Q4:

In a right angle triangle ABC, if ∠B = 90°,  ∠C = 60° and AB = 3cm then solve it.


Type: Long Difficulty: Easy

Show/Hide Answer
Answer: <p>Solution:</p> <p>Given,<br>&ang;B = 90&deg;, &ang;C = 60&deg; and AB = 3cm<br>Now,<br>or, &ang;A = &ang;B &minus; &ang;C<br>or, &ang;A = 90&deg;&minus; 60&deg;<br>&there4; &ang;A = 30&deg;<br>Again,<br>or, sin60&deg; = \(\frac{AB}{AC}\)<br>or, \(\frac{\sqrt{3}}{2}\) = \(\frac{3}{AC}\)<br>or, \(\sqrt{3}{AC}\) = 3 &times; 2<br>or, AC = \(\frac{3 &times; 2}{\sqrt{3}}\)<br>or, AC = \(\frac{\sqrt{3}&times; \sqrt{3} &times; 2}{\sqrt{3}}\)<br>&there4; AC = 2\(\sqrt{3}\)<br>Then.<br>or, sin30&deg; = \(\frac{p}{h}\)<br>or, \(\frac{1}{2}\) = \(\frac{BC}{2\sqrt{3}}\)<br>or, 2 BC = 2\(\sqrt{3}\)<br>or, BC = \(\frac{2\sqrt{3}}{2}\)<br>&there4; BC = \(\sqrt{3}\)<br>&there4; The value of AC is 2\(\sqrt{3}\) and BC is \(\sqrt{3}\).<br><br></p>

Q5:

Solve it, when ∠B = 90°,  ∠A = 30° and c = 5cm in a right angle triangle.


Type: Short Difficulty: Easy

Show/Hide Answer
Answer: <p>Solution:</p> <p>Given,<br>&ang;B = 90&deg;, &ang;A = 30&deg; and c = 5cm<br>or, &ang;C = &ang;B &minus; &ang;A<br>or, &ang;C = 90&deg;&minus;30&deg; <br>or, &ang;C = 60&deg; <br>Now,<br>or, sin60&deg; = \(\frac{p}{h}\)<br>or, \(\frac{\sqrt{3}}{2}\) = \(\frac{5}{h}\)<br>or, \(\sqrt{3}\) h = 10<br>&there4;h = \(\frac{10}{\sqrt{3}}\)<br>Again,<br>or, tan60&deg; = \(\frac{p}{b}\)<br>or, \(\sqrt{3}\) = \(\frac{5}{b}\)<br>or, \(\sqrt{3}\) b = 5<br>&there4; b = \(\frac{5}{\sqrt{3}}\)</p> <p></p>

Q6:

Solve when ∠B = 90°,  c = 2cm and a = 2cm in right angle triangle.


Type: Short Difficulty: Easy

Show/Hide Answer
Answer: <p>Solution:</p> <p>Given,<br>&ang;B = 90&deg;, c = 2cm and a = 2cm<br>Taking reference angle C<br>or, tanC = \(\frac{AB}{BC}\)<br>or, tanC = \(\frac{2}{2}\)<br>or, tanC = 1<br>or, tanC = tan45&deg;<br>&there4; C = 45&deg;<br>Now,<br>or, &ang;A = 90&deg;&minus; 45&deg;<br>&there4; &ang;A = 45&deg;<br>Again,<br>or, sin45&deg; = \(\frac{p}{h}\)<br>or, \(\frac{1}{\sqrt{2}}\) = \(\frac{BC}{AC}\)<br>or, \(\frac{1}{\sqrt{2}}\) = \(\frac{2}{AC}\)<br>&there4; AC = 2 \(\sqrt{2}\)cm</p>

Q7:

Solve it, when ∠A = 90°,  b = 6\(\sqrt{3}\)cm, and c = 6 cm in right angle triangle.


Type: Short Difficulty: Easy

Show/Hide Answer
Answer: <p>Solution:</p> <p>In right angle triangle ABC<br>&ang;A = 90&deg;, b = 6\(\sqrt{3}\)cm, and c = 6 cm<br>Now,<br>or, tanB = \(\frac{AC}{AB}\)<br>or, tanB = \(\frac{6 \sqrt{3}}{6}\)<br>or, tanB = \(\sqrt{3}\)<br>or, tanB = tan60&deg;<br>&there4; b = 60&deg;</p> <p>Taking reference&ang;C<br>or, sinC = \(\frac{AB}{BC}\)<br>or, sin30&deg; = \(\frac{6}{BC}\)<br>or, \(\frac{1}{2}\) = \(\frac{6}{BC}\)<br>&there4; BC = 12cm.</p>

Q8:

Solve when ∠B = 90°,  a = \(\sqrt{3}\) cm, and c = 1cm on right angle triangle.


Type: Short Difficulty: Easy

Show/Hide Answer
Answer: <p>Solution:</p> <p>In a right angle triangle <br>&ang;B = 90&deg;, a = \(\sqrt{3}\) cm and c = 1cm <br>Taking reference angle C<br>or, tanC = \(\frac{AB}{BC}\)<br>or, tanC = \(\frac{1}{\sqrt{3}}\)<br>or, tanC = tan30&deg;<br>&there4; C = 60&deg;<br>Now,<br>or, &ang;A = 90&deg;&minus; 30&deg;<br>or, &ang;A = 60&deg;</p> <p>Taking reference angle A<br>or, sinA = \(\frac{BC}{AC}\)<br>or, sin60&deg; =\(\frac{BC}{AC}\)<br>or, \(\frac{\sqrt{3}}{2}\) = \(\frac{\sqrt{3}}{AC}\)<br>or, \(\sqrt{3}\) AC = 2\(\sqrt{2}\)<br>or, AC = \(\frac{2\sqrt{3}}{\sqrt{3}}\)<br>&there4; AC = 2cm</p>

Q9:

Solve when ∠C = 90°,  ∠A = 30° and b = 20 cm in right angle triangle.


Type: Short Difficulty: Easy

Show/Hide Answer
Answer: <p>Solution:</p> <p>In right angle<br>&ang;C = 90&deg;, &ang;A = 30&deg; and b = 20 cm<br>Taking&ang;A as a reference<br>or, &ang;B = &ang;C&minus;&ang;A <br>or, &ang;B = 90&deg; &minus; 30&deg;<br>or, &ang;B = 60&deg;</p> <p>Taking &ang;B as a reference<br>or, sinB = \(\frac{p}{h}\)<br>or, sin60&deg; = \(\frac{20}{C}\)<br>or, \(\frac{\sqrt{3}}{2}\) = \(\frac{20}{C}\)<br>or, \(\sqrt{3}\) C = 40<br>or, C = \(\frac{40}{\sqrt{3}}\)<br>Then,<br>or, cos60&deg; = \(\frac{b}{h}\)<br>or, \(\frac{1}{2}\) = \(\frac{a}{\frac{40}{\sqrt{3}}}\)<br>or, \(\frac{1}{2}\) = \(\frac{a&times; \sqrt{3}}{40}\)<br>or, 40 = 2\(\sqrt{3}\) a<br>or, a = \(\frac{40}{2\sqrt{3}}\)<br>&there4; a = \(\frac{20}{\sqrt{3}}\)</p>

Q10:

In a right angle triangle \(\angle\)PQR, \(\angle\)Q = 90o , \(\angle\)R = \(\theta\) , PQ = 5cm, QR = 12cm, find all trignometric relation,


Type: Short Difficulty: Easy

Show/Hide Answer
Answer: <p>Solution:</p> <p>In right angle triangle \(\angle\)PQR,\(\angle\)Q = 90<sup>o</sup>, \(\angle\)R = \(\theta\) , PQ = 5cm, QR = 12cm</p> <p>Now,</p> <p>Hypotenuse(h) = ?</p> <p>Perpendicular(p) = 5cm</p> <p>Base(b) = 12cm</p> <p>We know, h = \(\sqrt{b^2-p^2}\)</p> <p>=\(\sqrt{12^2-5^2}\)</p> <p>=\(\sqrt{144-25}\)</p> <p>=\(\sqrt{169}\)</p> <p>= 13cm</p> <p>Now,</p> <p>sin\(\theta\) = \(\frac{p}{h}\) = \(\frac{5}{13}\)</p> <p>cos\(\theta\) =\(\frac{b}{h}\) =\(\frac{12}{13}\)</p> <p>tan\(\theta\) =\(\frac{p}{b}\) =\(\frac{5}{12}\)</p> <p>cosec\(\theta\) =\(\frac{h}{p}\) =\(\frac{13}{5}\)</p> <p>sec\(\theta\) =\(\frac{h}{b}\) =\(\frac{5}{13}\)</p> <p>cot\(\theta\) =\(\frac{b}{p}\) =\(\frac{12}{5}\)</p> <p></p>

Q11:

Show that the triangle having the sides 6cm, 8cm and 10cm is a right angled triangle.


Type: Very_short Difficulty: Easy

Show/Hide Answer
Answer: <p>Solution:</p> <p>Hypotenuse(h) = 10cm</p> <p>Perpendicular(p) = 8cm</p> <p>Base(b) = 6cm</p> <p>We know,</p> <p>h<sup>2</sup>= p<sup>2</sup>+ b<sup>2</sup></p> <p>or, 10<sup>2</sup>= 8<sup>2</sup>+ 6<sup>2</sup></p> <p>or, 100 = 64 + 36</p> <p>\(\therefore\) 100 = 100 (It satisfy pythagorus theorem)</p> <p>Hence, the given triangle is right angled triangle.</p> <p></p>

Q12:

Show that the triangle having the sides 12cm, 16cm and 20cm is a right angled triangle.


Type: Short Difficulty: Easy

Show/Hide Answer
Answer: <p>Solution:</p> <p>Hypotenuse(h) = 20cm</p> <p>Perpendicular(p) = 16cm</p> <p>Base(b) = 12cm</p> <p>We know,</p> <p>h<sup>2</sup>= p<sup>2</sup>+ b<sup>2</sup></p> <p>or, 20<sup>2</sup>= 16<sup>2</sup>+ 12<sup>2</sup></p> <p>or, 400 = 256 + 144</p> <p>\(\therefore\) 400 = 400 (It satisfy pythagorus theorem)</p> <p>Hence, the given triangle is right angled triangle.</p> <p></p>

Q13:

Show that the triangle having the sides 3cm, 4cm and 5cm is a right angled triangle.


Type: Very_short Difficulty: Easy

Show/Hide Answer
Answer: <p>Solution:</p> <p>Hypotenuse(h) = 5cm</p> <p>Perpendicular(p) = 4cm</p> <p>Base(b) = 3cm</p> <p>We know,</p> <p>h<sup>2</sup>= p<sup>2</sup>+ b<sup>2</sup></p> <p>or, 5<sup>2</sup>= 4<sup>2</sup>+ 3<sup>2</sup></p> <p>or, 25 = 16 + 9</p> <p>\(\therefore\) 25 = 25 (It satisfy pythagorus theorem)</p> <p>Hence, the given triangle is right angled triangle.</p> <p></p>

Q14:

In the adjoining figure ABC is a right angled triangle where <ABC = 90°. Express sinθ, cosθ and tanθ in terms of sides. 

.

Type: Short Difficulty: Easy

Show/Hide Answer
Answer: <p>Solution:</p> <p>Here,</p> <p>AB, BC and CA are perpendicular, base and hypotenuse respectively.</p> <p>Now,</p> <p>&there4; sin&theta; = \(\frac{p}{h}\) = \(\frac{AB}{AC}\)</p> <p>&there4; cos&theta; = \(\frac{b}{h}\) = \(\frac{BC}{AC}\)</p> <p>&there4; tan&theta; = \(\frac{p}{b}\) = \(\frac{AB}{BC}\)</p>

Q15:

In a right angled triangle \(\triangle\)ABC, right angle at B, AC = 10cm and sinA = \(\frac{4}{5}\). Find the length of AB.


Type: Short Difficulty: Easy

Show/Hide Answer
Answer: <p>Solution:</p> <p>In right angle triangle ABC, AC = 10cm</p> <p>sinA = \(\frac{BC}{AC}\)</p> <p>or, \(\frac{4}{5}\) = \(\frac{BC}{10}\)</p> <p>or, 5BC = 10 x 4</p> <p>or, BC = \(\frac{40}{5}\)</p> <p>\(\therefore\) BC = 8 cm</p> <p>By using pythagorus theorum</p> <p>or, AB<sup>2</sup>= AC<sup>2</sup> - BC<sup>2</sup></p> <p>or, AB = \(\sqrt{(10^2 - 8^2)}\)</p> <p>or, AB = \(\sqrt{100 - 64}\)</p> <p>or, AB = \(\sqrt{36}\)</p> <p>\(\therefore AB = 6cm</p> <p>Hence, the length of AB is 6cm.</p>

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Natural Resource and Human Dependency

Natural Resource and Human Dependency

Earth is a common home for humans and other animals. So, all living beings gets their needs from the earth. Living beings survives in water, air, forests, minerals and land. There is a relationship between natural resources and living beings.

Food

The most important need of every living being is food. In the absence of food survival of a living being is impossible. Humans cultivate on soil and extracts food from it. Humans cultivate wheat, maize, mustard and barley. Plants make food by photosynthesis. Humans are consumers as they fully depend on plants and other animals for food.

Habitat

Human lives on land. Humans live in the house, tiger lives in den, cows lives in the shed and so on. Humans use resources like land, cement, rock, soil, cement, sand, marble and so on to make their house.

Medicinal Plants

Medicinal plants are the vital components of pharmacy. Harro, barro, bhojo, sarpagandha, jatamasi (spikenard), panchaunle, yachagumba are important medicinal herbs. In Nepal, there are about 800 types of medicinal herbs in Nepal but only 70 are used as medicine. Forest degradation is the major cause of degradation in medicinal herbs.

Air

Air is an essential for living. Fresh air makes the man healthy. Humans take oxygen and give out carbon dioxide. Humans use wind energy in construction and transportation.

Water

Water is important natural resources. 70% of our body is composed of water. The scarcity of water has a negative effect on animals and human beings. Water is used to drink, clean, irrigate, bath and so on. Human settlement is successful on the bank of water resources. Industrial pollutants, drainage and wastage mixed in water resources pollutes the water.

Land

Most of the livings beings lives on land. Food and habitat of human and many animals depend on the land. Without land, humans are incapable of doing agriculture. Human settlements, unsustainable agricultural practices, investments in extractive resources, industry and tourism are causing serious changes in landscapes and ecosystems.

Lesson

Environment and its Balance

Subject

Science

Grade

Grade 8

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