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79 lines
2.2 KiB
Markdown
79 lines
2.2 KiB
Markdown
## Trigonometry
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### Question 1 a)
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It means to solve all missing/unknown angles and sidelengths. It can be achieved by using some of the following:
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1. Sine/cosine law
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2. Primary Trigonometry Ratios
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3. Similar / Congruent Triangle theorems
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4. Angle Theorems
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5. Pythagorean Theorem
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### Question 1 b)
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Draw a line bisector perpendiculr to $`\overline{XZ}`$. Then by using pythagorean theorem: $`7^2 - y^2 = h^2`$, where $`h`$ is the height.
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$`\therefore h = \sqrt{13} \approx 3.61cm`$
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### Question 1 c)
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We can draw a triangle $`ABC`$ where $`\angle A`$ is the angle between the hands, and $\overline{AB}`$ and $`\overline{BC}`$ are the long and short hands respectively.
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Since a clock is a circle, $`\angle A = \dfrac{360}{12} \times 2 = 60^o`$
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Let $`x`$ be the distance between the 2 hands. By using the law of cosines:
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$`x^2 = 12^2 + 15^2 - 2(15)(12)\cos60`$
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$`x = 13.7cm`$.
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The distance between the 2 hands is $`13.7cm`$.
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### Question 2 a)
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Lets split the tree into the 2 triangles shown on the diagram. By using the primary trigonmetry ratios, we know that the bottom triangle's height side lenghts that is part of
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the tree's height is $`100\tan 10`$, and $`100 \tan 25`$ for the top triangle.
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Therefore the tree's height is the sum of these 2 triangle's side length.
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Therefore the total height is $`100(\tan 25 + \tan 10) = 64.3`$
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The height of the tree is $`64.3m`$
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### Question 2 b)
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$`\angle G = 180 - 35 - 68 = 77 (ASTT) `$
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By using the law of sines.
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$`\overline{RG} = \dfrac{173.2 \sin 35}{\sin 77} = 102m`$
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By using the law of sines.
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$`\overline{TG} = \dfrac{173.2 \sin 68}{\sin 77} = 164.8m`$
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$`P = 173.2 + 164.8 + 102 = 440m`$
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The perimeter is $`440m`$
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### Question 2 c)
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We know the buildings must be on the same side because they both cast a shadow from the same one sun.
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Let the triangle formed by the flagpole be $`\triangle FPS`$ and the one by the building $`\triangle TBS`$
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$`\because \angle B = \angle P`$ (given)
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$`\because \angle S`$ is common.
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$`\therefore \triangle TBS \sim \triangle FPS`$ (AA similarity theorem)
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$`\dfrac{TB}{BS} = \dfrac{FP}{PS} \implies \dfrac{TB}{26} = \dfrac{25}{10]`$
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$`\therefore TB = \dfrac{25(26)}{10} = 65`$
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Therefore the building is $`65m`$ tall.
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### Question 3 a)
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