Bazı Trigonometri Problemleri-Çözümlü

Çözüm:

12-16-20 dik üçgendir:

${\displaystyle m\left(DAE\right)=\alpha \text{ve}\left|DE\right|=k\text{diyelim}}$

${\displaystyle \cos 2\alpha =\frac{\cancel{16}}{\cancel{20}}}$

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

${\displaystyle \frac{\left|DE\right|}{\left|AD\right|}=\sin \alpha }$

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

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

${\displaystyle \Rightarrow \left|AD\right|=\sqrt{10}\left|DE\right|}$

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

${\displaystyle \text{Pisagor teoremi ile}}$

${\displaystyle \left|AD\right|=\sqrt{10-1}k}$

${\displaystyle \times \times =3k}$

${\displaystyle \frac{k}{\left|EC\right|}=\tan 2\alpha }$

${\displaystyle \frac{k}{\left|EC\right|}=\frac{3}{4}}$

${\displaystyle \left|EC\right|=\frac{4k}{3}}$

${\displaystyle \frac{\left|BD\right|}{\left|DC\right|}=\frac{3k}{\frac{4k}{3}}}$

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

Çözüm:

Birinci yol:

${\displaystyle m\left(EDA\right)=x\text{olsun}}$

${\displaystyle m\left(BDF\right)=180-90-m\left(DBF\right)}$

${\displaystyle \times \times \times =40}$

${\displaystyle \text{Açıortayın kollara uzaklığı eşittir.}}$

${\displaystyle \text{O halde}CDF\text{ve}CDE\text{eşit üçgenlerdir.}}$

${\displaystyle \Rightarrow \left|CE\right|=\left|CF\right|=m\text{diyelim}}$

${\displaystyle \left|BC\right|-\left|AD\right|=\left|AC\right|}$

${\displaystyle \left|BF\right|+\left|CF\right|-\left|AD\right|=\left|CE\right|-\left|AE\right|}$

${\displaystyle \Rightarrow \left|BF\right|-\left|AD\right|=-\left|AE\right|}$

${\displaystyle \Rightarrow \left|AD\right|-\left|AE\right|=\left|BF\right|}$

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

${\displaystyle \Rightarrow \sec x-\tan x=\tan 40}$

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

${\displaystyle \frac{{\left(\sec x\right)}^2-{\left(\tan x\right)}^2}{\sec x-\tan x}=\cot 40}$

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

${\displaystyle \sec x+\tan x=\cot 40}$

Bunu ilk denklem ile toplarsak:

${\displaystyle \sec x-\tan x+\sec x+\tan x=\tan 40+\cot 40}$

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

${\displaystyle \frac{2}{\cos x}=\frac{{\left(\sin 40\right)}^2+{\left(\cos 40\right)}^2}{\sin 40\cdot \cos 40}}$

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

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

${\displaystyle \sin (90-x)=\sin 80}$

${\displaystyle x=10}$

${\displaystyle m\left(ADF\right)=180-40}$

${\displaystyle \Rightarrow m\left(CDF\right)+m\left(CDE\right)-10=140}$

${\displaystyle \Rightarrow 2m\left(CDF\right)=150}$

${\displaystyle m\left(CDF\right)=75}$

${\displaystyle \Rightarrow m\left(BDC\right)=40+75}$

${\displaystyle \times \times \times =115}$


İkinci yol:

${\displaystyle 2m\left(CDF\right)+40>180}$

${\displaystyle 2m\left(CDF\right)>140}$

${\displaystyle m\left(CDF\right)>70}$

${\displaystyle m\left(CDF\right)+40>110}$

${\displaystyle \Rightarrow m(BDC>110}$

${\displaystyle m\left(BDC\right)\text{ancak 115 olabilir.} }$

Çözüm:

${\displaystyle }$ ${\displaystyle m\left(DAC\right)=\alpha \text{diyelim}}$

${\displaystyle m\left(DAE\right)=m\left(BAC\right)=60}$

${\displaystyle \Rightarrow m\left(CAE\right)+\alpha =m\left(BAD\right)+\alpha }$

${\displaystyle \Rightarrow m\left(CAE\right)=m\left(BAD\right)}$

${\displaystyle \left|AB\right|=\left|AC\right|\text{ve}\left|AD\right|=\left|AE\right|}$

${\displaystyle \text{KAK teoremine göre}}$

${\displaystyle ABD=ACE}$

${\displaystyle \Rightarrow \left|BD\right|=2}$

${\displaystyle \Rightarrow \left|BC\right|=2+1}$

${\displaystyle \times \times x=3}$

${\displaystyle \left|AH\right|=\frac{3\sqrt{3}}{2}}$

${\displaystyle \left|DH\right|=\frac{3}{2}-1}$

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

${\displaystyle \left|ED\right|=\left|AD\right|}$

${\displaystyle \text{Pisagor teoremi ile}}$

${\displaystyle \left|ED\right|=\sqrt{{\left(\frac{1}{2}\right)}^2+{\left(\frac{3\sqrt{3}}{2}\right)}^2}}$

${\displaystyle \times x=\sqrt{7}}$


 

${\displaystyle m\left(BAE\right)=60-20}$

${\displaystyle \times \times \times =40}$

${\displaystyle m\left(DEA\right)=\frac{180-100}{2}}$

${\displaystyle \times \times \times =40}$

${\displaystyle m\left(ADE\right)=180-80}$

${\displaystyle \times \times \times =100}$

${\displaystyle {\left|AE\right|}^2=1+1-2\cos 100\text{(Kosinüs teoremi)}}$

${\displaystyle {\left|AE\right|}^2=2+2\cos 80}$

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

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

${\displaystyle \Rightarrow \left|AE\right|=2\cos 40}$