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

Trigonometry

Question 1 a)

It means to solve all missing/unknown angles and sidelengths. It can be achieved by using some of the following:

1. Sine/cosine law
2. Primary Trigonometry Ratios
3. Similar / Congruent Triangle theorems
4. Angle Theorems
5. Pythagorean Theorem

Question 1 b)

Draw a line bisector perpendiculr to \(`\overline{XZ}`\). Then by using pythagorean theorem: \(`7^2 - y^2 = h^2`\), where \(`h`\) is the height.

\(`\therefore h = \sqrt{13} \approx 3.61cm`\)

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.

Since a clock is a circle, \(`\angle A = \dfrac{360}{12} \times 2 = 60^o`\)

Let \(`x`\) be the distance between the 2 hands. By using the law of cosines:

\(`x^2 = 12^2 + 15^2 - 2(15)(12)\cos60`\)

\(`x = 13.7cm`\).

The distance between the 2 hands is \(`13.7cm`\).

Question 2 a)

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 the tree’s height is \(`100\tan 10`\), and \(`100 \tan 25`\) for the top triangle.

Therefore the tree’s height is the sum of these 2 triangle’s side length.

Therefore the total height is \(`100(\tan 25 + \tan 10) = 64.3`\)

The height of the tree is \(`64.3m`\)

Question 2 b)

\(`\angle G = 180 - 35 - 68 = 77 (ASTT) `\)

By using the law of sines.

\(`\overline{RG} = \dfrac{173.2 \sin 35}{\sin 77} = 102m`\)

By using the law of sines.

\(`\overline{TG} = \dfrac{173.2 \sin 68}{\sin 77} = 164.8m`\)

\(`P = 173.2 + 164.8 + 102 = 440m`\)

The perimeter is \(`440m`\)

Question 2 c)

We know the buildings must be on the same side because they both cast a shadow from the same one sun.

Let the triangle formed by the flagpole be \(`\triangle FPS`\) and the one by the building \(`\triangle TBS`\)

\(`\because \angle B = \angle P`\) (given)

\(`\because \angle S`\) is common.

\(`\therefore \triangle TBS \sim \triangle FPS`\) (AA similarity theorem)

\(`\dfrac{TB}{BS} = \dfrac{FP}{PS} \implies \dfrac{TB}{26} = \dfrac{25}{10]`\)

\(`\therefore TB = \dfrac{25(26)}{10} = 65`\)

Therefore the building is \(`65m`\) tall.

Question 3 a)