Hey, fellow mathletes! Welcome to your one-stop-shop for a solid practice session.
Feeling a bit rusty on your laws of exponents? Do massive numbers in scientific notation make your head spin? Or maybe you’re gearing up to conquer those classic right-angled triangle problems? You’ve come to the right place! Today, we’re diving into a mixed review designed to sharpen your skills in some core IB DP Math topics: Number & Algebra and Geometry & Trigonometry. We’ll start with the basics to build your confidence, then ramp it up with some IB-style questions. Let’s get to it!
Part 1: The Warm-Up – Nailing the Fundamentals
Let’s start by shaking off the dust. These questions are all about reinforcing the core rules and procedures. Don’t skip these – a strong foundation is everything!
Laws of Exponents & Scientific Notation
Question 1: Simplify the expression $(3a^4b^{-2})^3$.
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Solution: We distribute the exponent of 3 to each part inside the parentheses. Remember to multiply the exponents!
$$ (3a^4b^{-2})^3 = 3^3 \cdot (a^4)^3 \cdot (b^{-2})^3 = 27a^{4 \times 3}b^{-2 \times 3} = 27a^{12}b^{-6} $$
Or, if you need to express it with positive exponents: $$ \frac{27a^{12}}{b^6} $$
Question 2: Simplify the expression $$\frac{24x^8y^3}{8x^{-2}y^5}$$.
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Solution: We handle the coefficients, the $x$ terms, and the $y$ terms separately. When dividing powers with the same base, you subtract the exponents.
$$ \frac{24x^8y^3}{8x^{-2}y^5} = \frac{24}{8} \cdot x^{8 – (-2)} \cdot y^{3 – 5} = 3x^{10}y^{-2} = \frac{3x^{10}}{y^2} $$
Question 3: Write the number 607,500,000,000 in scientific notation.
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Solution: Scientific notation requires a single non-zero digit before the decimal point. We need to move the decimal point 11 places to the left.
$$ 607,500,000,000 = 6.075 \times 10^{11} $$
Question 4: Convert $8.41 \times 10^{-6}$ to standard notation.
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Solution: The exponent is -6, so we need to move the decimal point 6 places to the left, adding zeros as placeholders.
$$ 8.41 \times 10^{-6} = 0.00000841 $$
Question 5: A number is written as $345.1 \times 10^4$. Express this number in proper scientific notation ($a \times 10^k$, where $1 \le a < 10$ and $k$ is an integer).
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Solution: First, we need to adjust the coefficient, 345.1, so it’s between 1 and 10. We do this by moving the decimal point two places to the left: $345.1 = 3.451 \times 10^2$.
Now substitute this back into the original expression:
$$ (3.451 \times 10^2) \times 10^4 = 3.451 \times 10^{2+4} = 3.451 \times 10^6 $$
Right-Angled Triangles: Pythagoras & Trigonometry
Question 6: In a right-angled triangle, the two shorter sides have lengths of 9 cm and 12 cm. What is the length of the hypotenuse?
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Solution: Using the Pythagorean theorem, $a^2 + b^2 = c^2$.
$$ 9^2 + 12^2 = c^2 $$
$$ 81 + 144 = c^2 $$
$$ 225 = c^2 $$
$$ c = \sqrt{225} = 15 \text{ cm} $$
Question 7: A right-angled triangle has a hypotenuse of length 26 m and one leg of length 10 m. Find the length of the other leg.
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Solution: Again, using $a^2 + b^2 = c^2$. Let the unknown leg be $b$.
$$ 10^2 + b^2 = 26^2 $$
$$ 100 + b^2 = 676 $$
$$ b^2 = 676 – 100 = 576 $$
$$ b = \sqrt{576} = 24 \text{ m} $$
Question 8: In a right-angled triangle, the angle $\theta$ has an opposite side of 14 cm and an adjacent side of 20 cm. Find the size of angle $\theta$ to the nearest degree.
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Solution: We have the Opposite and Adjacent sides, so we use the tangent ratio (SOH CAH TOA).
$$ \tan(\theta) = \frac{\text{Opposite}}{\text{Adjacent}} = \frac{14}{20} = 0.7 $$
To find the angle, we use the inverse tangent function:
$$ \theta = \arctan(0.7) \approx 34.99^\circ $$
To the nearest degree, $\theta = 35^\circ$.
Question 9: From a point 40 meters away from the base of a flagpole, the angle of elevation to the top of the pole is $38^\circ$. Find the height of the flagpole to one decimal place.
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Solution: Let $h$ be the height of the flagpole. The height is the side opposite the angle, and the distance from the base is the adjacent side. We use the tangent ratio.
$$ \tan(38^\circ) = \frac{h}{40} $$
$$ h = 40 \times \tan(38^\circ) \approx 31.251 \text{ m} $$
To one decimal place, the height is 31.3 m.
Part 2: Gearing Up – IB Exam Style Section A Questions
Alright, let’s step it up a notch. These questions are similar in style to the short-form questions you’ll find in Section A of your IB exam. They might require a couple of steps to solve. You got this!
IB Exam Style Question
Question 10: The approximate volume of Jupiter is $1.43 \times 10^{15} \text{ km}^3$, and the approximate volume of Earth is $1.08 \times 10^{12} \text{ km}^3$.
(a) Find the combined volume of Jupiter and Earth. Give your answer in the form $a \times 10^k$, where $1 \le a < 10$ and $k \in \mathbb{Z}$.
(b) How many times greater is the volume of Jupiter than the volume of Earth? Give your answer to the nearest whole number.
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Solution:
(a) To add numbers in scientific notation, we need to make the exponents the same. Let’s convert Earth’s volume to have an exponent of 15.
$$ 1.08 \times 10^{12} = 0.00108 \times 10^{15} $$
Now we can add:
$$ (1.43 \times 10^{15}) + (0.00108 \times 10^{15}) = (1.43 + 0.00108) \times 10^{15} = 1.43108 \times 10^{15} \text{ km}^3 $$
This is already in the correct form. To 3 s.f., it’s $1.43 \times 10^{15} \text{ km}^3$.
(b) To find how many times greater, we divide the volume of Jupiter by the volume of Earth.
$$ \frac{1.43 \times 10^{15}}{1.08 \times 10^{12}} = \frac{1.43}{1.08} \times 10^{15-12} \approx 1.324 \times 10^3 = 1324 $$
To the nearest whole number, Jupiter is about 1324 times greater in volume than Earth.
IB Exam Style Question
Question 11: An isosceles triangle ABC has AB = AC = 15 cm and BC = 18 cm. M is the midpoint of BC.
(a) Find the length of AM.
(b) Find the size of angle ABC, giving your answer to one decimal place.
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Solution:
(a) Since M is the midpoint of BC, BM = 9 cm. Triangle ABM is a right-angled triangle with hypotenuse AB = 15 cm. We can use Pythagoras’ theorem to find AM.
$$ (\text{AM})^2 + (\text{BM})^2 = (\text{AB})^2 $$
$$ (\text{AM})^2 + 9^2 = 15^2 $$
$$ (\text{AM})^2 + 81 = 225 $$
$$ (\text{AM})^2 = 144 \implies \text{AM} = 12 \text{ cm} $$
(b) In the right-angled triangle ABM, we want to find angle ABC (which is the same as angle ABM). We have the adjacent side (BM = 9) and the hypotenuse (AB = 15).
$$ \cos(\text{angle ABC}) = \frac{\text{Adjacent}}{\text{Hypotenuse}} = \frac{9}{15} = 0.6 $$
$$ \text{angle ABC} = \arccos(0.6) \approx 53.13^\circ $$
To one decimal place, the angle is 53.1°.
IB Exam Style Question
Question 12: A park ranger in a tower 50 meters high spots a fire at an angle of depression of $7^\circ$. A second fire is spotted at an angle of depression of $3^\circ$. Both fires are in the same straight line from the base of the tower.
(a) Calculate the horizontal distance from the tower to the closer fire.
(b) Calculate the distance between the two fires. Give your answer to the nearest meter.
Tip
Remember that the angle of depression from the tower is equal to the angle of elevation from the ground. Drawing a diagram here is a lifesaver!
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Solution:
(a) Let the closer fire be F1. The angle of elevation from F1 to the top of the tower is $7^\circ$. The height of the tower is 50 m (opposite side). Let the distance be $d_1$ (adjacent side).
$$ \tan(7^\circ) = \frac{50}{d_1} \implies d_1 = \frac{50}{\tan(7^\circ)} \approx 407.22 \text{ m} $$
The distance to the closer fire is 407.2 m (to 1 d.p.).
(b) Let the farther fire be F2. The angle of elevation from F2 is $3^\circ$. Let its distance be $d_2$.
$$ \tan(3^\circ) = \frac{50}{d_2} \implies d_2 = \frac{50}{\tan(3^\circ)} \approx 954.05 \text{ m} $$
The distance between the fires is the difference between $d_2$ and $d_1$.
$$ \text{Distance} = d_2 – d_1 = 954.05 – 407.22 = 546.83 \text{ m} $$
To the nearest meter, the distance is 547 m.
Part 3: The Main Event – IB Exam Style Section B Questions
This is where it all comes together. Section B questions are longer, multi-part problems that test your ability to connect different concepts. They often build on each other, so read carefully and show your work. Take a deep breath, you can do this!
IB Exam Style Question
Question 13: A surveyor is trying to determine the height of a vertical cliff. The surveyor stands at point A and measures the angle of elevation to the top of the cliff, T, to be $32^\circ$. The surveyor then walks 100 meters directly towards the base of the cliff to a new point, B, and measures the angle of elevation to T again. This time it is $48^\circ$. The points A, B, and the base of the cliff, C, are on the same horizontal level.
(a) Draw a clear, labelled diagram of the situation.
(b) Let $h$ be the height of the cliff (TC) and let $x$ be the distance from B to C. Write down two trigonometric equations, one involving the triangle ATC and the other involving the triangle BTC.
(c) By solving the two equations simultaneously, show that $h = \frac{100 \tan(32^\circ) \tan(48^\circ)}{\tan(48^\circ) – \tan(32^\circ)}$.
(d) Hence, find the height of the cliff, $h$, to the nearest meter.
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Solution:
(a) Your diagram should show two right-angled triangles, ATC and BTC, sharing the same height $h = TC$. The distance AC = $100 + x$, and BC = $x$. Angle TAC = $32^\circ$ and angle TBC = $48^\circ$.
(b) In triangle BTC: $$ \tan(48^\circ) = \frac{h}{x} $$ In triangle ATC: $$ \tan(32^\circ) = \frac{h}{100+x} $$
(c) First, rearrange both equations to make $x$ the subject.
From the first equation: $x = \frac{h}{\tan(48^\circ)}$
From the second equation: $100+x = \frac{h}{\tan(32^\circ)} \implies x = \frac{h}{\tan(32^\circ)} – 100$
Now, set the two expressions for $x$ equal to each other:
$$ \frac{h}{\tan(48^\circ)} = \frac{h}{\tan(32^\circ)} – 100 $$
Now, we solve for $h$. Get all terms with $h$ on one side:
$$ 100 = \frac{h}{\tan(32^\circ)} – \frac{h}{\tan(48^\circ)} $$
Factor out $h$:
$$ 100 = h \left( \frac{1}{\tan(32^\circ)} – \frac{1}{\tan(48^\circ)} \right) $$
Combine the fractions inside the parenthesis:
$$ 100 = h \left( \frac{\tan(48^\circ) – \tan(32^\circ)}{\tan(32^\circ)\tan(48^\circ)} \right) $$
Finally, isolate $h$ by multiplying by the reciprocal of the fraction:
$$ h = \frac{100 \tan(32^\circ) \tan(48^\circ)}{\tan(48^\circ) – \tan(32^\circ)} $$
(d) Now we just need to calculate the value:
$$ h = \frac{100(0.6248…)(1.1106…)}{1.1106… – 0.6248…} = \frac{69.39…}{0.4857…} \approx 142.85 \text{ m} $$
To the nearest meter, the height of the cliff is 143 m.
IB Exam Style Question
Question 14: A rectangular box has dimensions $L = 5 \times 10^3$ cm, $W = 4 \times 10^3$ cm, and $H = 2 \times 10^3$ cm.
(a) Calculate the volume of the box. Give your answer in scientific notation.
(b) A space diagonal, $d$, is a line connecting two opposite corners of the box (e.g., from the bottom-front-left corner to the top-back-right corner). The length of the space diagonal is given by the formula $d = \sqrt{L^2 + W^2 + H^2}$. Calculate the length of the space diagonal. Give your answer to 3 significant figures.
(c) Let $\theta$ be the angle the space diagonal makes with the bottom face of the box. Calculate $\theta$ to the nearest degree.
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Solution:
(a) Volume $V = L \times W \times H$.
$$ V = (5 \times 10^3) \times (4 \times 10^3) \times (2 \times 10^3) $$
$$ V = (5 \times 4 \times 2) \times (10^3 \times 10^3 \times 10^3) = 40 \times 10^{3+3+3} = 40 \times 10^9 $$
In proper scientific notation, this is $4.0 \times 10^{10} \text{ cm}^3$.
(b) We use the formula $d = \sqrt{L^2 + W^2 + H^2}$.
$$ L^2 = (5 \times 10^3)^2 = 25 \times 10^6 $$
$$ W^2 = (4 \times 10^3)^2 = 16 \times 10^6 $$
$$ H^2 = (2 \times 10^3)^2 = 4 \times 10^6 $$
$$ d = \sqrt{25 \times 10^6 + 16 \times 10^6 + 4 \times 10^6} = \sqrt{(25+16+4) \times 10^6} = \sqrt{45 \times 10^6} $$
$$ d = \sqrt{45} \times \sqrt{10^6} = \sqrt{45} \times 10^3 \approx 6.708 \times 10^3 \text{ cm} $$
To 3 significant figures, $d = 6.71 \times 10^3$ cm.
(c) To find the angle $\theta$ with the bottom face, we can form a right-angled triangle. The ‘opposite’ side is the height of the box, $H$. The ‘adjacent’ side is the diagonal of the bottom face of the box. Let’s call this bottom diagonal $d_{base}$.
First find $d_{base}$ using Pythagoras on the bottom face: $d_{base} = \sqrt{L^2 + W^2} = \sqrt{25 \times 10^6 + 16 \times 10^6} = \sqrt{41 \times 10^6} = \sqrt{41} \times 10^3$.
Now, consider the right-angled triangle formed by $H$, $d_{base}$, and the space diagonal $d$. The angle $\theta$ is between $d_{base}$ and $d$.
$$ \tan(\theta) = \frac{\text{Opposite}}{\text{Adjacent}} = \frac{H}{d_{base}} = \frac{2 \times 10^3}{\sqrt{41} \times 10^3} = \frac{2}{\sqrt{41}} $$
$$ \tan(\theta) \approx 0.3123… $$
$$ \theta = \arctan(0.3123…) \approx 17.35^\circ $$
To the nearest degree, $\theta = 17^\circ$.
That’s a Wrap!
And there you have it! A solid workout across some fundamental IB Math concepts. How did you do? Remember, the goal of practice isn’t to be perfect every time, but to identify where you’re strong and where you need a bit more focus. The more you practice connecting these ideas, the more comfortable you’ll feel when you see them on an exam. Keep up the amazing work!
