Chapter 8: Introduction to Trigonometry

A fire engine has a telescopic ladder that can be extended. The ladder leans against a vertical wall of a building. The foot of the ladder is 6 m away from the base of the wall, and the top of the ladder reaches a window 8 m above the ground.

1. What is the length of the ladder?
   (A) 10 m

   (B) 12 m

   (C) 14 m

   (D) 9 m
   Show Solution
   Solution: (A) 10 m
   Step 1: Let the length of the ladder be $AC$. Base $BC = 6$ m, Height $AB = 8$ m.
   Step 2: By Pythagoras theorem, $AC^2 = AB^2 + BC^2 = 8^2 + 6^2 = 64 + 36 = 100$.
   Step 3: $AC = \sqrt{100} = 10$ m.
2. If $\theta$ is the angle made by the ladder with the ground, find $\cos \theta$.
   (A) 3/5

   (B) 4/5

   (C) 3/4

   (D) 4/3
   Show Solution
   Solution: (A) 3/5
   Step 1: $\cos \theta = \frac{\text{Base}}{\text{Hypotenuse}} = \frac{BC}{AC}$.
   Step 2: $\cos \theta = \frac{6}{10} = \frac{3}{5}$.

A slide in a children's park is designed such that its length is 10 m. The slide makes an angle of $30^\circ$ with the ground.

1. What is the height of the top of the slide from the ground?
   (A) 5 m

   (B) $5\sqrt{3}$ m

   (C) 10 m

   (D) $10\sqrt{3}$ m
   Show Solution
   Solution: (A) 5 m
   Step 1: Let height be $h$. Length of slide (hypotenuse) $= 10$ m. Angle $= 30^\circ$.
   Step 2: $\sin 30^\circ = \frac{\text{Opposite}}{\text{Hypotenuse}} = \frac{h}{10}$.
   Step 3: $\frac{1}{2} = \frac{h}{10} \Rightarrow h = 5$ m.
2. If the slide was steeper, making an angle of $60^\circ$ with the ground, what would be the height?
   (A) 5 m

   (B) $5\sqrt{3}$ m

   (C) 10 m

   (D) $10\sqrt{3}$ m
   Show Solution
   Solution: (B) $5\sqrt{3}$ m
   Step 1: $\sin 60^\circ = \frac{h}{10}$.
   Step 2: $\frac{\sqrt{3}}{2} = \frac{h}{10} \Rightarrow h = 5\sqrt{3}$ m.

In a right-angled triangle $ABC$, right-angled at $B$, it is given that $\tan A = \frac{4}{3}$.

1. Find the value of $\sin A$.
   (A) 3/5

   (B) 4/5

   (C) 3/4

   (D) 5/4
   Show Solution
   Solution: (B) 4/5
   Step 1: $\tan A = \frac{\text{Opp}}{\text{Adj}} = \frac{4}{3}$. Let Opp $= 4k$, Adj $= 3k$.
   Step 2: Hypotenuse $= \sqrt{(4k)^2 + (3k)^2} = \sqrt{16k^2 + 9k^2} = \sqrt{25k^2} = 5k$.
   Step 3: $\sin A = \frac{\text{Opp}}{\text{Hyp}} = \frac{4k}{5k} = \frac{4}{5}$.
2. Evaluate $\frac{1 - \sin A}{1 + \cos A}$.
   (A) 1/8

   (B) 1/4

   (C) 1/5

   (D) 1/2
   Show Solution
   Solution: (A) 1/8
   Step 1: $\sin A = 4/5$. $\cos A = \frac{\text{Adj}}{\text{Hyp}} = \frac{3}{5}$.
   Step 2: Numerator: $1 - 4/5 = 1/5$. Denominator: $1 + 3/5 = 8/5$.
   Step 3: Ratio: $\frac{1/5}{8/5} = \frac{1}{8}$.