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179 lines
5.0 KiB
Markdown
179 lines
5.0 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
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Let the triangle be $`triangle ABC`$, with $`AB = 1500`$ and $`BC = 4000`$. Let $`\angle A = \alpha`$ Then the angle of depression = $`\theta = 90 - \alpha`$ (CAT)
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$`\tan(\alpha) = \dfrac{4000}{1500}`$
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$`\alpha = \tan^{-1} (\dfrac{4000}{1500}) = 69.4`$
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$`\therefore \theta = 90 - \alpha = 20.6`$
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Therefore her angle of depression is $`20.6^o`$.
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By pythagorean theorem, we know that $`\overline{AC}^2 = \overline{AB}^2 + \overline{BC}^2 \implies \overline{AC}^2 = 1500^2 + 4000^2 \implies \overline{AC} = 4272m`$.
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She flew about $`4272m`$ before touching down.
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### Question 4 a)
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Since they form a right triangle, we know the distance between them is the adjacent side to the $`55^o`$ angle. We also know that the hypotenuse is $`45m`$ long.
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Therefore let $`x`$ be the distance. $`x = 45 \cos 55 = 25.8m`$.
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They are $`25.8m`$ apart.
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### Question 4 b)
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Lets draw a 3D tetrahedron. Let $`KA = 75m, \angle KAB = 31, \angle BAK = 54, \angle BKT = 61`$. And $`K`$ be Ken's position, $`A`$ be Adam's position, $`B`$ be the base of the cliff, and
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$`T`$ be the top of the cliff.
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We first need to find $`KB`$, where then we can use primary trigonmetry ratios to find out $`BT`$, which is the height.
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$`\angle B = 180 - 54 - 31 = 95`$ (ASTT)
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By using the law of sines.
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$`\dfrac{75}{\sin 95} = \dfrac{KB}{\sin A}`$
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$`KB = \dfrac{75\sin 31}{\sin95} = 38.78`$
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$`BT = 38.78 \tan 61 = 70m`$
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The height of the cliff is 70m.
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### Question 4 c)
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For congruency:
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SSS; When all 3 corresponding sides are the same
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SAS; when 2 corresponding sides are the same, and one corresponding angle is the same.
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ASA; when 2 corresponding angles are the same, and one corresponding side is the same.
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For Similarity:
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RRR; When the ratio of the 3 corresponding sides are all the same.
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RAR: when the ratio of 2 of the 3 corresponding sides are all the same and one corresponding angle is the same.
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AA When 2 corresponding angles are the same.
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### Question 5 a)
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The sinelaw is a relation between a triangles side length and the sine of its corresponding angle, which this relation has the same ratio as all
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the other sides with their corresponding angles.
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You can use sinelaw when you have a oblique tiangle and you have at least 2 sides/angles and 1 angle/side.
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### Question 5 b)
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$`\because b\sin A \lt a \lt b`$
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$`\because 5.3 \lt 7.5 \lt 9.3`$
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There are 2 cases.
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Let $`B^\prime`$ be the other possible point of $`B`$.
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#### Case 1
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$`\angle B = \sin^{-1}(\dfrac{5.3}{7.2}) = 47.4`$
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$`\angle C = 180 - 35 - 47.4 = 97.6`$ (ASTT)
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By using the law of cosines:
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$`\overline{AB}^2 = 9.3^2 + 7.2^2 - 2(9.3)(7.2)\cos97.6 \implies \overline{AB} = 12.49mm`$
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#### Case 2
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$`\angle B^\prime = 180 - \angle B = 132.6`$ (SAT)
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$`\angle C = 180 - 132.6 - 35 = 12.4`$ (ASTT)
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By using the law of cosines.
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$`\overline{AB}^2 = 9.3^2 + 7.2^2 - 2(9.3)(7.2)\cos12.4 \implies \overline{AB} = 2.74 mm`$
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