Image_20261104204427_157_128.jpg

S3 · 中三 questions. Answer in your own words, then submit to get a score.

1. In $\triangle ABC$ , $A = 50^\circ$ and $B = 25^\circ$ . Explain why $BC \neq 2AC$ .
2. In $\triangle ABC$ , $B = 30^\circ$ , $C = 20^\circ$ , and side $a = 9$ cm (i.e., $BC = 9$ cm). Find angle $A$ and explain how to find side $b$ (i.e., $AC$ ).
3. In $\triangle ABC$ , $B = 72^\circ$ , $C = 48^\circ$ , and side $b = 13$ cm (i.e., $AC = 13$ cm). Calculate angle $A$ and side $c$ (i.e., $AB$ ), showing your steps.
4. In a triangle with sides 26 cm and 42 cm, and the angle opposite the 42 cm side is $64^\circ$ , find the angle $\theta$ opposite the 26 cm side. Explain why $\theta$ must be acute.
5. In $\triangle ABC$ , $A = 50^\circ$ , $B = 25^\circ$ , and $BC = 10$ cm. Is it possible that $AB > 13$ cm? Justify your answer using the sine formula.

Transcript

## Page 1 S5 Math CP Chapter 9 Applications of Trigonometry in 2-dimensional Problems Name ________________ Class __ ( ) 2 1. In

\triangle ABC

B = 30^\circ

C = 20^\circ

and

a = 9

cm. Solve the triangle. 2. In

\triangle ABC

B = 72^\circ

C = 48^\circ

and

b = 13

cm. Solve the triangle.

\hookrightarrow

Ex 9A: 7–9

\begin{aligned} & A + 30^\circ + 20^\circ = 180^\circ \quad (\text{sum of } \angle s) \\ & \therefore A = 130^\circ \\ \\ & \frac{b}{\sin 30^\circ} = \frac{9}{\sin 130^\circ} \\ & b = 5.87 \, \text{cm} \\ \\ & \frac{c}{\sin 20^\circ} = \frac{a}{\sin 130^\circ} \\ & c = 4.02 \, \text{cm} \end{aligned}

\begin{aligned} & A + 72^\circ + 48^\circ = 180^\circ \quad (\text{sum of } \angle s) \\ & \therefore A = 60^\circ \\ \\ & \frac{a}{\sin 60^\circ} = \frac{13}{\sin 72^\circ} \\ & a = 11.8 \, \text{cm} \\ \\ & \frac{c}{\sin 48^\circ} = \frac{13}{\sin 72^\circ} \\ & c = 10.2 \, \text{cm} \end{aligned}

3. In

\triangle ABC

A = 50^\circ

and

B = 25^\circ

. Explain (a) Is it true that

BC = 2AC

? Explain your answer. (b) If

BC = 10

cm, is it possible that

AB > 13

cm? Explain your answer.

\begin{aligned} & \angle C = 50^\circ + 25^\circ = 180^\circ \\ & \therefore \angle C = 105^\circ \\ \\ & \frac{BC}{\sin 50^\circ} = \frac{AC}{\sin 25^\circ} \\ & BC = 1.81 \, AC \\ & \therefore \text{False} \end{aligned}

\begin{aligned} & \frac{10}{\sin 25^\circ} = \frac{AB}{\sin 105^\circ} \\ & AB = 12.6 \, \text{cm} \\ & \therefore \text{No} \end{aligned}

**Instant Example 1** In the figure, find

\theta

. By the sine formula,

\frac{15 \, \text{cm}}{\sin \theta} = \frac{18 \, \text{cm}}{\sin 35^\circ}

\sin \theta = \frac{15 \sin 35^\circ}{18}

\theta = 28.6^\circ, \text{ corr. to 3 sig. fig.}

\because PQ < QR

\therefore \theta

must be an acute angle smaller than

35^\circ

. **Instant Practice 1** In the figure, find

\theta

. By the sine formula,

\frac{(26)}{\sin \theta} = \frac{(42)}{\sin (64^\circ)}

\sin \theta = \frac{(26) \sin (64^\circ)}{(42)}

\theta = 33.8^\circ, \text{ corr. to 3 sig. fig.}