Update Trigonometry.md

Grade 10/Math/MPM2DZ/Math Oral Presentation Questions/Trigonometry.md

@@ -18,7 +18,7 @@ $`\therefore h = \sqrt{13} \approx 3.61cm`$
 
 ### Question 1 c)
 
-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.
+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.
 
 Since a clock is a circle, $`\angle A = \dfrac{360}{12} \times 2 = 60^o`$
 
@@ -69,10 +69,110 @@ $`\because \angle S`$ is common.
 
 $`\therefore \triangle TBS \sim \triangle FPS`$ (AA similarity theorem)
 
-$`\dfrac{TB}{BS} = \dfrac{FP}{PS} \implies \dfrac{TB}{26} = \dfrac{25}{10]`$
+$`\dfrac{TB}{BS} = \dfrac{FP}{PS} \implies \dfrac{TB}{26} = \dfrac{25}{10}`$
 
 $`\therefore TB = \dfrac{25(26)}{10} = 65`$
 
 Therefore the building is $`65m`$ tall.
 
-### Question 3 a)
+### Question 3
+
+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)
+
+$`\tan(\alpha) = \dfrac{4000}{1500}`$
+
+$`\alpha = \tan^{-1} (\dfrac{4000}{1500}) = 69.4`$
+
+$`\therefore \theta = 90 - \alpha = 20.6`$
+
+Therefore her angle of depression is $`20.6^o`$.
+
+By pythagorean theorem, we know that $`\overline{AC}^2 = \overline{AB}^2 + \overline{BC}^2 \implies \overline{AC}^2 = 1500^2 + 4000^2 \implies \overline{AC} = 4272m`$. 
+
+She flew about $`4272m`$ before touching down.
+
+### Question 4 a)
+
+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.
+
+Therefore let $`x`$ be the distance. $`x = 45 \cos 55 = 25.8m`$.
+
+They are $`25.8m`$ apart.
+
+### Question 4 b)
+
+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
+$`T`$ be the top of the cliff.
+
+We first need to find $`KB`$, where then we can use primary trigonmetry ratios to find out $`BT`$, which is the height.
+
+$`\angle B = 180 - 54 - 31 = 95`$ (ASTT)
+
+By using the law of sines.
+
+$`\dfrac{75}{\sin 95} = \dfrac{KB}{\sin A}`$
+
+$`KB = \dfrac{75\sin 31}{\sin95} = 38.78`$
+
+$`BT = 38.78 \tan 61 = 70m`$
+
+The height of the cliff is 70m.
+
+
+### Question 4 c)
+
+For congruency:
+
+SSS; When all 3 corresponding sides are the same
+
+SAS; when 2 corresponding sides are the same, and one corresponding angle is the same.
+
+ASA; when 2 corresponding angles are the same, and one corresponding side is the same.
+
+For Similarity:
+
+RRR; When the ratio of the 3 corresponding sides are all the same.
+
+RAR: when the ratio of 2 of the 3 corresponding sides are all the same and one corresponding angle is the same.
+
+AA When 2 corresponding angles are the same.
+
+### Question 5 a)
+
+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 
+the other sides with their corresponding angles.
+
+You can use sinelaw when you have a oblique tiangle and you have at least 2 sides/angles and 1 angle/side.
+
+### Question 5 b)
+
+
+$`\because b\sin A \lt a \lt b`$
+
+$`\because 5.3 \lt 7.5 \lt 9.3`$
+
+There are 2 cases.
+
+Let $`B^\prime`$ be the other possible point of $`B`$.
+
+#### Case 1
+
+$`\angle B = \sin^{-1}(\dfrac{5.3}{7.2}) = 47.4`$
+
+$`\angle C = 180 - 35 - 47.4 = 97.6`$ (ASTT)
+
+By using the law of cosines:
+
+$`\overline{AB}^2 = 9.3^2 + 7.2^2 - 2(9.3)(7.2)\cos97.6 \implies \overline{AB} = 12.49mm`$
+
+#### Case 2
+
+$`\angle B^\prime = 180 - \angle B = 132.6`$ (SAT)
+
+$`\angle C = 180 - 132.6 - 35 = 12.4`$ (ASTT)
+
+By using the law of cosines.
+
+$`\overline{AB}^2 = 9.3^2 + 7.2^2 - 2(9.3)(7.2)\cos12.4 \implies \overline{AB} = 2.74 mm`$
+
+