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Two angles are complementary. One angle is 12° larger than three times the other angle. Find the measure of both angles.

Mathematics

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3 parts
3

19.5∘ na 70.5∘\text{19.5}^\circ \text{ na } 70.5^\circ

4

miaka 9\text{miaka 9}

miaka 36\text{miaka 36}

5

x=5 au x=−13x = 5 \text{ au } x = -\frac{1}{3}

Step-by-step solution

13 steps

Part 3· 4 steps

Milinganyo ya Pembe Kamilishi

  1. Fafanua vigezo na uandike milinganyo.

    Tuseme pembe mbili ni α\alpha na β\beta.
    Kwa kuwa pembe hizi ni kamilishi (complementary), jumla yao ni 90∘90^\circ.

    α+β=90∘(1)\alpha + \beta = 90^\circ \quad (1)

    Pembe moja ni 12∘12^\circ kubwa kuliko mara tatu ya pembe nyingine.

    α=3β+12∘(2)\alpha = 3\beta + 12^\circ \quad (2)
  2. Badilisha mlinganyo (2) kwenye mlinganyo (1).

    Badilisha 3β+12∘3\beta + 12^\circ kwa α\alpha katika mlinganyo (1):

    (3β+12∘)+β=90∘(3\beta + 12^\circ) + \beta = 90^\circ
  3. Tatua kwa β\beta.

    4β+12∘=90∘4\beta + 12^\circ = 90^\circ

    Toa 12∘12^\circ kutoka pande zote mbili:

    4β=90∘−12∘4\beta = 90^\circ - 12^\circ 4β=78∘4\beta = 78^\circ

    Gawanya kwa 4:

    β=78∘4\beta = \frac{78^\circ}{4} β=19.5∘\beta = 19.5^\circ
  4. Tafuta α\alpha.

    Tumia mlinganyo (2) kupata α\alpha:

    α=3(19.5∘)+12∘\alpha = 3(19.5^\circ) + 12^\circ α=58.5∘+12∘\alpha = 58.5^\circ + 12^\circ α=70.5∘\alpha = 70.5^\circ

    Pembe hizo ni 19.5∘ na 70.5∘\boxed{\text{19.5}^\circ \text{ na } 70.5^\circ}.

Part 4

miaka 9\text{miaka 9}

Kuunda na Kutatua Milinganyo ya Aljebra

  1. Fafanua vigezo vya umri wa sasa.

    Tuseme umri wa sasa wa mwana ni SS na umri wa sasa wa mama ni MM.

  2. Unda milinganyo kulingana na taarifa.

    "Mama ana umri mara 4 ya mwanawe":

    M=4S(1)M = 4S \quad (1)

    "Katika miaka 5 ijayo, jumla ya umri wao itakuwa miaka 55."
    Umri wa mwana baada ya miaka 5 utakuwa S+5S+5.
    Umri wa mama baada ya miaka 5 utakuwa M+5M+5.
    Jumla ya umri wao baada ya miaka 5:

    (S+5)+(M+5)=55(S+5) + (M+5) = 55

    Rahisisha mlinganyo wa pili:

    S+M+10=55S + M + 10 = 55 S+M=55−10S + M = 55 - 10 S+M=45(2)S + M = 45 \quad (2)
  3. Tatua mfumo wa milinganyo.

    Badilisha mlinganyo (1) (M=4SM=4S) kwenye mlinganyo (2):

    S+4S=45S + 4S = 45 5S=455S = 45

    Gawanya kwa 5:

    S=455S = \frac{45}{5} S=9S = 9

    Umri wa sasa wa mwana ni miaka 9.

  4. Tafuta umri wa mama.

    Tumia mlinganyo (1) kupata MM:

    M=4SM = 4S M=4(9)M = 4(9) M=36M = 36

    Umri wa sasa wa mama ni miaka 36.
    Umri wao wa sasa ni: Mwana miaka 9\boxed{\text{miaka 9}}, Mama miaka 36\boxed{\text{miaka 36}}.

Part 5

x=5 au x=−13x = 5 \text{ au } x = -\frac{1}{3}

Kutatua Milinganyo ya Quadratic kwa Njia ya Factorization

  1. Andika mlinganyo na utambue vigawo.

    Mlinganyo ni 3x2−14x−5=03x^2 - 14x - 5 = 0.
    Hapa, a=3a=3, b=−14b=-14, c=−5c=-5.

  2. Tafuta namba mbili zinazozidisha acac na kujumlisha bb.

    ac=3×(−5)=−15ac = 3 \times (-5) = -15.
    Tunahitaji namba mbili ambazo zinazidisha -15 na kujumlisha -14. Namba hizo ni -15 na 1.
    (−15×1=−15-15 \times 1 = -15 na −15+1=−14-15 + 1 = -14).

  3. Andika upya muda wa kati (bxbx) kwa kutumia namba hizi.

    Badilisha −14x-14x na −15x+x-15x + x:

    3x2−15x+x−5=03x^2 - 15x + x - 5 = 0
  4. Factor kwa kupanga (factor by grouping).

    Panga maneno mawili ya kwanza na mawili ya mwisho:

    (3x2−15x)+(x−5)=0(3x^2 - 15x) + (x - 5) = 0

    Toa kigawo cha kawaida kutoka kila kundi:

    3x(x−5)+1(x−5)=03x(x - 5) + 1(x - 5) = 0

    Toa kigawo cha kawaida cha binomial (x−5)(x-5):

    (x−5)(3x+1)=0(x - 5)(3x + 1) = 0
  5. Weka kila kigawo sawa na sifuri na utatue kwa xx.

    x−5=0au3x+1=0x - 5 = 0 \quad \text{au} \quad 3x + 1 = 0 x=5au3x=−1x = 5 \quad \text{au} \quad 3x = -1 x=5aux=−13x = 5 \quad \text{au} \quad x = -\frac{1}{3}

    Thamani za xx ni x=5 au x=−13\boxed{x = 5 \text{ au } x = -\frac{1}{3}}.

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