IB myp 4-5 MATHEMATICS – Practice Questions- All Topics
Topic :Trigonometry–Sine rule and cosine rule
Topic :Trigonometry- Weightage : 21 %
All Questions for Topic : Converse of Pythagoras’ theorem,Sine rule and cosine rule, including applications (link to trigonometric functions)
Question (a)
▶️Answer/Explanation
Ans:
Expression: \(\cos \theta \tan \theta\)
1. Recall that \(\tan \theta = \frac{\sin \theta}{\cos \theta}\).
2. Substituting this value into the expression, we get: \(\cos \theta \cdot \frac{\sin \theta}{\cos \theta}\).
3. The \(\cos \theta\) terms cancel out, leaving us with just \(\sin \theta\).
So, \(\cos \theta \tan \theta = \sin \theta\).
Now, let’s choose the correct option from the draggable items:
- \(1\)
- \(\frac{\cos ^2 \theta}{\sin \theta}\)
- \(\sin \theta\)
The correct option is: \(\sin \theta\)
Question (b)
▶️Answer/Explanation
Ans:
Expression: \(1 – \sin^2 \theta + \cos^2 \theta\)
1. Recall the Pythagorean identity for sine and cosine: \(\sin^2 \theta + \cos^2 \theta = 1\).
2. Substitute this identity into the expression: \(1 – \sin^2 \theta + \cos^2 \theta = 1 – 1\).
3. \(1 – 1\) simplifies to \(0\).
So, \(1 – \sin^2 \theta + \cos^2 \theta = 0\).
Now, let’s choose the correct option from the draggable items:
- \(0\)
- \(2 \sin^2 \theta\)
- \(2 \cos^2 \theta\)
The correct option is: \(0\)
Question
Given that $B A C$ and $A C B$ are angles in radians. Show that the length of side $B C$ is $16.33 \mathrm{~cm}$ to the nearest two decimal places.
▶️Answer/Explanation
Ans:
The given equation involves trigonometric ratios and a proportion. Let’s solve for the value of \(BC\).
The trigonometric ratios are:
\(\sin 45^\circ = \frac{1}{\sqrt{2}}\) (since \(\sin 45^\circ = \cos 45^\circ = \frac{1}{\sqrt{2}}\))
\(\sin 60^\circ = \frac{\sqrt{3}}{2}\)
The equation is:
\(\frac{\sin 45^\circ}{BC} = \frac{\sin 60^\circ}{20 \, \text{cm}}\)
Substitute the trigonometric values:
\(\frac{\frac{1}{\sqrt{2}}}{BC} = \frac{\frac{\sqrt{3}}{2}}{20 \, \text{cm}}\)
Now, cross-multiply to solve for \(BC\):
\(BC \cdot \frac{1}{\sqrt{2}} = \frac{\sqrt{3}}{2} \cdot 20 \, \text{cm}\)
Simplify:
\(BC = \frac{\sqrt{3}}{\sqrt{2}} \cdot 20 \, \text{cm} = \sqrt{\frac{3}{2}} \cdot 20 \, \text{cm} = 10 \sqrt{6} \, \text{cm}\)
So, the value of \(BC\) is \(10 \sqrt{6}\) cm.