Ngā Tauira Pātai e Matapaki ana i ngā Mahi Pānga-toru
He wāhanga nui ngā mahi pākoki i roto i te pāngarau, e kitea pinepine ana i roto i ngā momo mara pūtaiao, tae atu ki te ahupūngao, te hangarau, me te pūtaiao rorohiko. I roto i tēnei tuhinga, ka matapakihia e mātou ētahi tauira rapanga, ā, ka whakaratohia he matapakinga hōhonu mō ngā mahi pākoki ine. Mā te mārama ki ēnei tauira, ko te tumanako ka whakapakari ake ngā kaipānui i tō rātou māramatanga me te kaha ki te whakaoti rapanga e pā ana ki ngā mahi pākoki ine.
He Kupu Whakataki ki ngā Mahi Pāngatoru
Ko ngā mahi whārite pātoru e tino kitea ana ko te sine (sin), te cosine (cos), me te tangent (tan). He mahi nui tā ēnei mahi e toru i roto i te whanaungatanga i waenga i ngā koki me ngā roa i roto i ngā tapatoru tika, tae atu ki ngā ngaru me ngā wiri.
Ngā Tātai Taketake:
1. Sine (hara)
\[
\sin(\theta) = \frac{\text{opposite}}{\text{hypotenuse}}
\]
2. Kosine (cos)
\[
\cos(\theta) = \frac{\text{adjacent}}{\text{hypotenuse}}
\]
3. Pānga (pango)
\[
\tan(\theta) = \frac{\text{opposite}}{\text{adjacent}}
\]
Ngā Tuakiri Pātoru
– Pythagoras:
\[
\sin^2(\theta) + \cos^2(\theta) = 1
\]
– Te whakataurite i te pānihi ki te sine me te cosine:
\[
tan(\theta) = \frac{\sin(\theta)}{\cos(\theta)}
\]
– Tuakiri tāpiri:
\[
\sin(2\theta) = 2\sin(\theta)\cos(\theta)
\]
\[
\cos(2\theta) = \cos^2(\theta) – \sin^2(\theta)
\]
Me titiro tātou ki ētahi tauira pātai me tētahi kōrero hōhonu ake.
Tauira Pātai 1: Te Tatau i te Uara o ngā Mahi Pānga-toru i tētahi Koki
Pātai:
Tātaihia ngā uara o te sin(30°), te cos(45°), me te tan(60°).
Kōrero:
E ai ki te ripanga o ngā uara pākoki taketake, kei a tātou:
– \(\sin(30°) = \frac{1}{2} = 0.5\)
– \(\cos(45°) = \frac{\sqrt{2}}{2} \approx 0.707\)
– \(\tan(60°) = \sqrt{3} \approx 1.732\)
Ko ngā uara e toru i runga ake nei he uara pākoki e whakamahia pinepinetia ana, ā, he mea pai kia maumaharahia ēnei nā te mea he maha ngā wā ka puta mai i roto i ngā pātai.
Tauira Pātai 2: Te Tatau i ngā Koki mā te Whakamahi i ngā Mahi Pāngatoru Whakamuri
Pātai:
Mena ko te \(\sin(\theta) = 0.5\), whakatauhia te uara o \(\theta\).
Kōrero:
Hei kimi i te uara o \(\theta\), me whakamahi tātou i te mahi whakamuri o te sine, arā, \(\arcsin\) me \(\sin^{-1}\ rānei.
\[
\theta = \sin^{-1}(0.5)
\]
I roto i te wā [0°, 360°], ko ngā uara e rite ana o \(\theta\) ko:
\[
\theta = 30° \text{ me } 150°
\]
nā te mea ko \(\sin(30°) = 0.5\) me \(\sin(150°) = 0.5\). Nō reira, ko ngā uara koki e rua e tutuki ana ko te 30° me te 150°.
Tauira Pātai 3: Te Whakamahi i ngā Tuakiri Pātoru
Pātai:
Whakamātauria ngā tuakiri pākoki
\[
\sin^2(\theta) + \cos^2(\theta) = 1.
\]
Kōrero:
I ahu mai tēnei tuakiri i te ariā Pythagorean i roto i ngā tapatoru matau. Me kī he tapatoru matau kei reira he koki \(\theta\), he taha whakarara \(a\), he taha tata \(b\), me te hypotenuse \(c\). Kātahi,
\[
a^2 + b^2 = c^2.
\]
Ki te wehea e tātou ngā taha e rua mā te \(c^2\), ka puta:
\[
\left(\frac{a}{c}\right)^2 + \left(\frac{b}{c}\right)^2 = 1.
\]
Na te mea
\[
\sin(\theta) = \frac{a}{c} \quad \text{and} \quad \cos(\theta) = \frac{b}{c},
\]
nā reira,
\[
\sin^2(\theta) + \cos^2(\theta) = 1.
\]
Koinei te huarahi e whakaatu ai tātou i tēnei tuakiri.
Tauira Pātai 4: Te Whakamahi i ngā Āhuatanga Pāngatoru hei Whakaoti i ngā Tapatoru
Pātai:
E hoatu ana te tapatoru ABC me te koki A 45°, te koki B 60°, me te taha AB 10 cm te roa. Kimihia te roa o ngā taha AC me BC.
Kōrero:
Whakamahia te ture sine hei kimi i te roa o ngā taha AC me BC.
\[
\frac{a}{\sin(A)} = \frac{b}{\sin(B)} = \frac{c}{\sin(C)}
\]
Tuatahi, ka kitea e tātou te koki C:
\[
C = 180° – A – B = 180° – 45° – 60° = 75°.
\]
Mēnā ko AB = 10 cm, \(A = 45°\), me \(B = 60°\), ka taea e tātou te whakamahi i te ture sine:
\[
\frac{AC}{\sin(60°)} = \frac{10}{\sin(75°)}.
\]
\[
AC = \frac{10 \sin(60°)}{\sin(75°)}.
\]
\[
AC = \frac{10 \times \frac{\sqrt{3}}{2}}{\sin(75°)} = \frac{10 \times \frac{\sqrt{3}}{2}}{\cos(15°)}.
\]
E mōhio ana tātou ko \(\cos(15°) = \cos(45° – 30°) = \cos 45° \cos 30° + \sin 45° \sin 30° = \frac{1}{\sqrt{2}} \cdot \frac{\sqrt{3}}{2} + \frac{1}{\sqrt{2}} \cdot \frac{1}{2}\).
\[
\cos(15°) = \frac{\sqrt{6} + \sqrt{2}}{4}.
\]
Nō reira:
\[
AC = \frac{10 \times \frac{\sqrt{3}}{2}}{\frac{\sqrt{6} + \sqrt{2}}{4}} = \frac{10 \times 2\sqrt{3}}{\sqrt{6} + \sqrt{2}} \approx 10.39 \text{ cm}.
\]
Waihoki, ka kitea e tātou a BC:
\[
\frac{BC}{\sin(45°)} = \frac{10}{\sin(75°)}.
\]
\[
BC = \frac{10 \sin(45°)}{\sin(75°)} \approx 8.66 \text{ cm}.
\]
I te mutunga o tēnei tuhinga, kua matapakihia e mātou ētahi tauira raruraru me ā rātou matapakinga e pā ana ki ngā mahi whārite. Mā te mahi tonu me te māramatanga hōhonu ki ngā tātai taketake, ngā tuakiri whārite, me ā rātou whakamahinga i roto i ngā tapatoru, e tumanakohia ana ka pai ake te mōhio o ngā kaipānui ki tēnei rauemi. He taputapu nui ngā mahi whārite, ehara i te mea i roto i te pāngarau anake engari i roto hoki i ngā momo kaupapa e whakawhirinaki ana ki te tātari i ngā koki me ngā roa. Ko te tumanako he tohutoro whai hua tēnei tuhinga mā ngā kaipānui.
