Tambayoyi Misali Game da Da'irori da Tangents
Da'ira muhimmin batu ne a fannin lissafi, inda ake nuna zurfafan ra'ayoyi game da nisa, kusurwoyi, da siffa. Wani ra'ayi da ake yawan tattaunawa a kai a wannan batu shine layin tangent zuwa da'ira. A cikin wannan labarin, za mu tattauna misalai da dama na matsaloli da suka shafi da'ira da tangent.
Fahimtar Asali Game da Da'irori da Tangents
Da'ira
Da'ira siffa ce ta geometric da aka samar ta hanyar saitin dukkan maki a cikin jirgin sama wanda ke da tazara mai tsayi daga wani wuri da aka ba da ake kira tsakiyar da'irar. Wannan tazara mai tsayi ana kiranta da radius na da'irar.
Tangent
Layin da ke taɓa da'ira shine layi da ke taɓa da'irar a daidai lokaci ɗaya. Ana kiran wannan wurin da wurin da ke haɗa da'ira. Tangents suna da wasu muhimman halaye, waɗanda suka haɗa da:
– Layin tangent koyaushe yana daidai da radius na da'irar a wurin tangent.
– Tsawon tangent daga wani wuri a wajen da'irar zuwa da'irar iri ɗaya ne idan aka zana tangent guda biyu daga wannan wurin.
Tambayoyi da Tattaunawar Samfura
A ƙasa za mu gabatar da wasu misalai na tambayoyi waɗanda suka tattauna manufar da'irori da tangent dalla-dalla.
Misali Tambaya ta 1: Nemo Tsawon Layin Tangent
Tambaya:
An ba da da'ira mai tsakiya \(O\) da radius \(r = 6 \, \text{cm}\). Daga wurin \(P\) a wajen da'irar wanda yake nisan santimita 10 daga tsakiyar da'irar, an zana tangent guda biyu \(PA\) da \(PB\) zuwa da'irar. Lissafa tsawon tangent \(PA\).
Tattaunawa:
A cikin wannan matsalar, za mu iya amfani da ka'idar Pythagorean. Zana alwatika \(\triangle OAP\):
– \(OP = 10 \, \text{cm}\) (nisa daga wurin waje zuwa tsakiyar da'irar)
– \(OA = 6 \, \text{cm}\) (radius na da'irar)
– \(PA\) shine layin tangent da dole ne a samo
\[
OP^2 = OA^2 + PA^2
\]
\[
10^2 = 6^2 + PA^2
\]
\[
100 = 36 + PA^2
\]
\[
PA^2 = 64
\]
\[
PA = \sqrt{64} = 8 \, \rubutu{cm}
\]
Don haka, tsawon layin tangent \(PA\) shine 8 cm.
Misali Tambaya ta 2: Gano Ma'anar Daidaitawa
Tambaya:
An ba da da'ira mai lissafin \((x – 3)^2 + (y – 4)^2 = 25\) da layi \(y = 2x + 1\). A tantance ma'aunin daidaito tsakanin da'irar da layin.
Tattaunawa:
Da farko, mun gano tsakiya da radius na da'irar:
– Tsakiya \(O(3, 4)\)
– Radius \(r = \sqrt{25} = 5\)
Domin gano ma'anar tangency, bari mu ɗauka cewa ma'anar tangency ita ce \(T(x_1, y_1)\) wanda shi ma yana kan layin \(y = 2x + 1\). Sannan:
\[
y_1 = 2x_1 + 1
\]
\(T(x_1, y_1)\) dole ne ya cika lissafin da'irar:
\[
(x_1 – 3)^2 + (y_1 – 4)^2 = 25
\]
Sauya \(y_1 = 2x_1 + 1\) cikin lissafin da'ira:
\[
(x_1 – 3)^2 + ((2x_1 + 1) – 4)^2 = 25
\]
\[
(x_1 – 3)^2 + (2x_1 – 3)^2 = 25
\]
Muna buƙatar ƙididdige murabba'i biyu.
\[
(x_1 – 3)^2 = x_1^2 – 6x_1 + 9
\]
\[
(2x_1 – 3)^2 = 4x_1^2 – 12x_1 + 9
\]
Haɗa sakamakon biyu:
\[
x_1^2 – 6x_1 + 9 + 4x_1^2 – 12x_1 + 9 = 25
\]
\[
5x_1^2 – 18x_1 + 18 = 25
\]
Cire 25 daga ɓangarorin biyu:
\[
5x_1^2 – 18x_1 – 7 = 0
\]
Warware lissafin murabba'i:
\[
x_1 = \frac{18 \pm \sqrt{18^2 + 4 \sau 5 \sau 7}}{2 \sau 5}
\]
\[
x_1 = \frac{18 \pm \sqrt{324 + 140}}{10}
\]
\[
x_1 = \frac{18 \pm \sqrt{464}}{10}
\]
\[
x_1 = \frac{18 \pm 2\sqrt{116}}{10}
\]
\[
x_1 = \frac{18 \pm 2\sqrt{4 \sau 29}}{10}
\]
\[
x_1 = \frac{18 \pm 4\sqrt{29}}{10}
\]
\[
x_1 = 1.8 \pm 0.4 \ sqrt{29}
\]
Lissafa ƙimar \(y_1\):
Wanda ya gamsar da y = 2x + 1:
– Idan \(x_1 = 1.8 + 0.4\sqrt{29}\), to \(y_1 = 2(1.8 + 0.4\sqrt{29}) + 1\)
– Idan \(x_1 = 1.8 – 0.4\sqrt{29}\), to \(y_1 = 2(1.8 – 0.4\sqrt{29}) + 1\)
Kimantawa:
Don haka mun sami maki biyu na haɗuwar lissafin da'irar tare da wannan layin.
Misali Tambaya ta 3: Tantance Daidaito na Layin Tangent
Tambaya:
An ba da da'ira mai lissafin \((x – 2)^2 + (y – 3)^2 = 20\). Kayyade lissafin layin tangent zuwa da'irar da ta ratsa ta wurin \(6, 7)\).
Tattaunawa:
Ana iya samun tangent ɗin da'ira mai tsakiya \((h, k)\) da radius \(r\) daga wani wuri na waje da aka sani ta hanyar lissafi:
Layin tangent yana ratsawa ta wurin waje \((x_1, y_1)\):
\[
(x - 2) (x_1 - 2) + (y - 3) (y_1 - 3) = 20
\]
Maye gurbin wurin waje \(6, 7)\):
\[
(x - 2) (6 - 2) + (y - 3) (7 - 3) = 20
\]
\[
4(x – 2) + 4(y – 3) = 20
\]
\[
4(x – 2 + y – 3) = 20
\]
\[
4x + 2y -20 = 20
\]
\[
4x + 4y -20 = 20
\]
\[
x + y = 5
\]
Daidaiton layin tangent shine:
\[
x + y = 9
\]
Don haka, bambancin lissafin layin ta hanyar ma'aunin layin da'ira yana da girma sosai kuma yana iya canzawa dangane da sakamakon ko wakilcin gani.
Kammalawa
Tattaunawar da'irori da tangent ta ƙunshi fannoni da dama na asali na lissafi, tun daga amfani da dabarun asali kamar ka'idar Pythagorean zuwa warware daidaiton quadratic. Ta hanyar waɗannan misalan, za mu iya haɓaka fahimtar yadda ake amfani da waɗannan ra'ayoyi a cikin yanayi mai rikitarwa. Da fatan, wannan labarin ya taimaka wajen samar da cikakken hoto na yadda ake tunkarar da kuma magance matsalolin da suka shafi da'irori da tangent.
