Year11 MATH 2-2-4 Warm-up Questions-Applied Trigonometry (General Mathematics)

To master Unit 2 of General Mathematics, you must move beyond the right-angled trigonometry of earlier years. Chapter 4 introduces the Sine Rule, the Cosine Rule, and the Area Rule, which allow you to solve for any triangle, anywhere. This skill is critical for advanced navigation, surveying, and 3D architectural modeling.

Concepts and Skills Covered:

1. The Sine Rule: Using $\frac{a}{\sin A} = \frac{b}{\sin B}$ to find missing sides and angles in non-right-angled triangles.
2. The Cosine Rule: Applying $a^2 = b^2 + c^2 – 2bc \cos A$ when provided with two sides and an included angle (SAS) or three sides (SSS).
3. Area of a Triangle: Calculating area using $\text{Area} = \frac{1}{2}ab \sin C$.
4. Strategic Selection: Determining which trigonometric rule is most efficient based on the given information.
5. Navigation and Bearings: Integrating bearings into trigonometric diagrams.

Q1. In $\Delta ABC$, you are given $\angle A = 40^{\circ}$, $\angle B = 60^{\circ}$, and side $b = 10\text{ cm}$. Which calculation correctly finds the length of side $a$?

A. $a = 10 \times \sin 40^{\circ} \times \sin 60^{\circ}$

B. $a = \frac{10 \sin 40^{\circ}}{\sin 60^{\circ}}$

C. $a = \frac{10 \sin 60^{\circ}}{\sin 40^{\circ}}$

D. $a^2 = 10^2 + \sin 40^{\circ} – 2(10)\cos 60^{\circ}$

Hint: Recall that the Sine Rule relates side lengths to the sine of the angle directly across from them.

Q2. To find a missing angle in a triangle where all three side lengths are known, which trigonometric rule is the most direct to use?

A. Sine Rule

B. Pythagoras’ Theorem

C. SOH CAH TOA

D. Cosine Rule

Hint: Consider which formula allows you to input three sides to output one angle.

Q3. In $\Delta PQR$, $p = 7\text{ cm}$, $q = 9\text{ cm}$, and $\angle R = 48^{\circ}$. Which formula would find the length of side $r$ ?

A. $r = \sqrt{7^2 + 9^2}$

B. $r^2 = 7^2 + 9^2 – 2(7)(9)\cos 48^{\circ}$

C. $\frac{r}{\sin 48^{\circ}} = \frac{7}{9}$

D. $r = 7 \cos 48^{\circ} + 9 \sin 48^{\circ}$

Hint: When you have two sides and the angle trapped between them, the Cosine Rule is the appropriate tool.

Q4. Calculate the area of $\Delta XYZ$ if $x = 10\text{ cm}$, $y = 15\text{ cm}$, and $\angle Z = 30^{\circ}$.

A. $64.9\text{ cm}^2$

B. $75\text{ cm}^2$

C. $37.5\text{ cm}^2$

D. $150\text{ cm}^2$

Hint: The formula for the area of a non-right-angled triangle involves sine and the included angle.

Q5. When using the Sine Rule to find an angle, which condition might lead to the ‘ambiguous case’ (two possible triangles)?

A. When given two angles and the included side (ASA).

B. When given two sides and a non-included acute angle (SSA).

C. When the triangle is right-angled.

D. When given all three side lengths (SSS).

Ambiguity arises when a specific set of side and angle information could theoretically draw two different shapes.

Q6. A surveyor needs to find the distance between two points, A and B, separated by a lake. They stand at point C and measure $AC = 120\text{ m}$, $BC = 150\text{ m}$, and $\angle ACB = 40^{\circ}$. What is the first step?

A. Calculate the average of the two known sides.

B. Use the Sine Rule to find ∠ABC.

C. Use the Cosine Rule to find the distance AB.

D. Assume the triangle is right-angled at C.

Hint: Evaluate the information provided: two sides and the angle between them.

Q7. If $\cos A$ is calculated to be a negative value (e.g., $-0.25$) when using the Cosine Rule, what does this tell you about $\angle A$?

A. Angle A is acute ($0^{\circ} < A < 90^{\circ}$)..

B. Angle A is obtuse ($90^{\circ} < A < 180^{\circ}$)..

C. The triangle cannot exist.

D. The calculation is incorrect.

Hint: Recall the behaviour of the cosine graph or the Unit Circle for angles greater than $90$ degrees.

Q8. A ship travels on a bearing of $060^{\circ}$ for $20\text{ km}$, then turns and travels on a bearing of $150^{\circ}$ for $15\text{ km}$. To find the direct distance back to the start, what is the internal angle at the turn?

A. ∠=60∘

B. ∠=150∘

C. ∠=210∘

D. ∠=90∘

Hint: Draw a diagram with ‘North’ lines at each turn to visualize the relationship between the bearings.

Q9. Which of these is the correct rearrangement of the Cosine Rule ($a^2 = b^2 + c^2 – 2bc \cos A$) to solve for the angle $A$?

A. $\cos A = \frac{2bc}{b^2 + c^2 – a^2}$

B. $\cos A = \frac{a^2 + b^2 – c^2}{2ab}$

C. $\cos A = \frac{b^2 + c^2 – a^2}{2bc}$

D. $\cos A = a^2 – b^2 – c^2 – 2bc$

Hint: Isolate the term containing ‘cos A’ on one side of the equation first.

Q10. A $10\text{ m}$ ladder leans against a wall. If the ladder makes an angle of $70^{\circ}$ with the ground, how high up the wall does it reach?

A. 3.42 m

B. 9.40 m

C. 27.47 m

D. 10.64 m

Hint: Identify the ladder as the hypotenuse and the height as the opposite side in a right-angled triangle.

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一般数学ユニット2をマスターするには、これまでの直角三角法の枠を超えなければなりません。第4章では、正弦定理、余弦定理、面積定理を紹介します。これらの定理は、あらゆる三角形を、あらゆる場所で解くのに役立ちます。このスキルは、高度なナビゲーション、測量、3D建築モデリングに不可欠です。

学習内容とスキル：

正弦定理：$\frac{a}{\sin A} = \frac{b}{\sin B}$ を用いて、直角でない三角形の欠けている辺と角度を求めます。

余弦定理：2辺と1つの夾角（SAS）または3辺（SSS）が与えられている場合、$a^2 = b^2 + c^2 – 2bc \cos A$ を適用します。

三角形の面積：$\text{面積} = \frac{1}{2}ab \sin C$ を用いて面積を計算します。

戦略的選択：与えられた情報に基づいて、どの三角関数の規則が最も効率的かを決定する。

ナビゲーションと方位：方位を三角関数図に組み込む。

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Right answers

Q1. B. $a = \frac{10 \sin 40^{\circ}}{\sin 60^{\circ}}$

Applying the Sine Rule, we set up the ratio $\frac{a}{\sin A} = \frac{b}{\sin B}$ and rearrange to solve for $a$.

Q2. D. Cosine Rule

The Cosine Rule can be rearranged to isolate the cosine of an angle, making it the standard choice for ‘Side-Side-Side’ (SSS) scenarios.

Q3. B. $r^2 = 7^2 + 9^2 – 2(7)(9)\cos 48^{\circ}$

This is the correct application of the Cosine Rule for the Side-Angle-Side (SAS) configuration.

Q4. C. $37.5\text{ cm}^2$

Using $\text{Area} = \frac{1}{2}xy \sin Z$, we calculate $0.5 \times 10 \times 15 \times \sin 30^{\circ} = 37.5$.

Q5. B. When given two sides and a non-included acute angle (SSA).

The ambiguous case occurs because the sine of an angle $\theta$ is the same as the sine of its supplement ($180^{\circ} – \theta$).

Q6. C. Use the Cosine Rule to find the distance $AB$.

Since the surveyor has two sides and the included angle (SAS), the Cosine Rule is the only way to find the third side directly.

Q7. B. Angle A is obtuse ($90^{\circ} < A < 180^{\circ}$).

The cosine function is negative in the second quadrant, which corresponds to obtuse angles in a triangle.

Q8. D. $\angle = 90^{\circ}$

The difference between the bearings $150^{\circ}$ and the back-bearing of the first leg ($60^{\circ} + 180^{\circ} = 240^{\circ}$) or simple geometry shows the internal angle is $90^{\circ}$.

Q9. C. $\cos A = \frac{b^2 + c^2 – a^2}{2bc}$

By moving $2bc \cos A$ to the left and $a^2$ to the right, then dividing by $2bc$, we isolate the cosine of the angle.

Q10. B. $9.40\text{ m}$

Using $h = 10 \sin 70^{\circ} \approx 9.40$. This is a basic right-angled application to contrast with the newer general rules.