Bazı Trigonometri Problemleri-Çözümlü

Çözüm:

12-16-20 dik üçgendir:

m (D A E) = α | D E | = k

\cos 2 α = \frac{16}{20}

\times \times = \frac{4}{5}

\frac{| D E |}{| A D |} = \sin α

\times \times = \sqrt{\frac{1 - \frac{4}{5}}{2}}

\times \times = \frac{1}{\sqrt{10}}

\Rightarrow | A D | = \sqrt{10} | D E |

\times \times \times = \sqrt{10} k

Pisagor teoremi ile

| A D | = \sqrt{10 - 1} k

\times \times = 3 k

\frac{k}{| E C |} = \tan 2 α

\frac{k}{| E C |} = \frac{3}{4}

| E C | = \frac{4 k}{3}

\frac{| B D |}{| D C |} = \frac{3 k}{\frac{4 k}{3}}

\times \times \times = \frac{9}{4}

Çözüm:

Birinci yol:

m (E D A) = x

m (B D F) = 180 - 90 - m (D B F)

\times \times \times = 40

Açıortayın kollara uzaklığı eşittir.

O halde C D F C D E

\Rightarrow | C E | = | C F | = m

| B C | - | A D | = | A C |

| B F | + | C F | - | A D | = | C E | - | A E |

\Rightarrow | B F | - | A D | = - | A E |

\Rightarrow | A D | - | A E | = | B F |

\Rightarrow \frac{m}{\cos x} - m \tan x = m \tan 40

\Rightarrow \sec x - \tan x = \tan 40

\frac{1}{\sec x - \tan x} = \frac{1}{\tan 40}

\frac{{(\sec x)}^{2} - {(\tan x)}^{2}}{\sec x - \tan x} = \cot 40

\frac{(\sec x + \tan x) \cdot (\sec x - \tan x)}{\sec x - \tan x} = \cot 40

\sec x + \tan x = \cot 40

Bunu ilk denklem ile toplarsak:

\sec x - \tan x + \sec x + \tan x = \tan 40 + \cot 40

2 \sec x = \frac{\sin 40}{\cos 40} + \frac{\cos 40}{\sin 40}

\frac{2}{\cos x} = \frac{{(\sin 40)}^{2} + {(\cos 40)}^{2}}{\sin 40 \cdot \cos 40}

\frac{2}{\cos x} = \frac{1}{\frac{1}{2} \cdot \sin 80}

\frac{2}{\sin (90 - x)} = \frac{2}{\sin 80}

\sin (90 - x) = \sin 80

x = 10

m (A D F) = 180 - 40

\Rightarrow m (C D F) + m (C D E) - 10 = 140

\Rightarrow 2 m (C D F) = 150

m (C D F) = 75

\Rightarrow m (B D C) = 40 + 75

\times \times \times = 115

İkinci yol:

2 m (C D F) + 40 > 180

2 m (C D F) > 140

m (C D F) > 70

m (C D F) + 40 > 110

\Rightarrow m (B D C > 110

m (B D C)

Çözüm:

m (D A C) = α

m (D A E) = m (B A C) = 60

\Rightarrow m (C A E) + α = m (B A D) + α

\Rightarrow m (C A E) = m (B A D)

| A B | = | A C | | A D | = | A E |

KAK teoremine göre

A B D = A C E

\Rightarrow | B D | = 2

\Rightarrow | B C | = 2 + 1

\times \times x = 3

| A H | = \frac{3 \sqrt{3}}{2}

| D H | = \frac{3}{2} - 1

\times x = \frac{1}{2}

| E D | = | A D |

Pisagor teoremi ile

| E D | = \sqrt{{(\frac{1}{2})}^{2} + {(\frac{3 \sqrt{3}}{2})}^{2}}

\times x = \sqrt{7}

m (B A E) = 60 - 20

\times \times \times = 40

m (D E A) = \frac{180 - 100}{2}

\times \times \times = 40

m (A D E) = 180 - 80

\times \times \times = 100

{| A E |}^{2} = 1 + 1 - 2 \cos 100

{| A E |}^{2} = 2 + 2 \cos 80

\times x = 2 + 2 (2 \cos^{2} 40 - 1)

\times x = 4 \cos^{2} 40

\Rightarrow | A E | = 2 \cos 40

Neoturanism