Basic Trigonometry Problems: Very Difficult Problems with Solutions

Solution:
[tex]\angle ABC[/tex] is obtuse, so [tex]D[/tex] lies on the extension of [tex]AB[/tex] beyond [tex]B[/tex]. [tex]\angle CBD=180^{\circ}-135^{\circ}=45^{\circ}[/tex], so in the right triangle [tex]BDC[/tex] we have [tex]BD=CD=x[/tex]. In the right triangle [tex]ADC[/tex]:
[tex]\tan 30^{\circ}=\frac{CD}{AD}=\frac{x}{24+x}\iff 24+x=\sqrt{3}x\iff x=\frac{24}{\sqrt{3}-1}=12\left(\sqrt{3}+1\right)[/tex]

Solution:
[tex]c=AB=\sqrt{4+8}=2\sqrt{3}[/tex]. Let [tex]AL[/tex] be the bisector of [tex]\alpha[/tex], [tex]L\in BC[/tex]. By the bisector theorem [tex]\frac{CL}{LB}=\frac{AC}{AB}=\frac{b}{c}[/tex], so [tex]CL=\frac{ab}{b+c}[/tex]. In the right triangle [tex]ACL[/tex]:
[tex]\tan \frac{\alpha}{2}=\frac{CL}{AC}=\frac{a}{b+c}=\frac{2}{2\sqrt{2}+2\sqrt{3}}=\frac{1}{\sqrt{3}+\sqrt{2}}=\sqrt{3}-\sqrt{2}[/tex]

Solution:
The legs are [tex]a=c\sin \alpha[/tex] and [tex]b=c\sin \beta=c\cos\alpha[/tex], because [tex]\beta=90^{\circ}-\alpha[/tex]. So [tex]S=\frac{ab}{2}=\frac{c^2}{2}\sin \alpha\cos\alpha[/tex]. Squaring the given equality:
[tex]m^2=\sin ^2\alpha+\cos^2\alpha+2\sin \alpha\cos\alpha=1+2\sin \alpha\cos\alpha[/tex]
Then [tex]\sin \alpha\cos\alpha=\frac{m^2-1}{2}[/tex] and [tex]S=\frac{c^2\left(m^2-1\right)}{4}[/tex].

Solution:
In the right triangle [tex]ABA_1[/tex]: [tex]BA_1=\sqrt{5^2-4^2}=3[/tex]. Let [tex]AC=BC=x[/tex]; then [tex]CA_1=x-3[/tex]. In the right triangle [tex]AA_1C[/tex]: [tex]x^2=4^2+(x-3)^2[/tex], so [tex]6x=25[/tex] and [tex]x=\frac{25}{6}[/tex]. Then [tex]CA_1=\frac{25}{6}-3=\frac{7}{6}\gt 0[/tex], so [tex]A_1[/tex] lies between [tex]B[/tex] and [tex]C[/tex]. Hence
[tex]\tan \gamma=\frac{AA_1}{CA_1}=\frac{4}{\frac{7}{6}}=\frac{24}{7}[/tex]

Solution:
In the right triangle [tex]AFC[/tex]: [tex]CF=AF\times \tan \alpha[/tex]. The line [tex]AH[/tex] is perpendicular to [tex]BC[/tex], so [tex]\angle HAF=90^{\circ}-\beta[/tex]. In the right triangle [tex]AFH[/tex]:
[tex]HF=AF\times \tan (90^{\circ}-\beta)=\frac{AF}{\tan \beta}[/tex]
From [tex]CF=2HF[/tex] we get [tex]AF\times \tan \alpha=\frac{2AF}{\tan \beta}[/tex], so [tex]\tan \alpha\times \tan \beta=2[/tex].

Solution:
[tex]\angle BAC=15^{\circ}[/tex]. Let [tex]CH[/tex] be the altitude; then [tex]\angle ACH=90^{\circ}-15^{\circ}=75^{\circ}[/tex]. Since [tex]MA=MC=5[/tex], [tex]\angle ACM=15^{\circ}[/tex] and [tex]\angle HCM=60^{\circ}[/tex]. Since [tex]\angle ACL=45^{\circ}[/tex], [tex]\angle HCL=30^{\circ}[/tex], so [tex]L[/tex] lies between [tex]H[/tex] and [tex]M[/tex]. In the right triangles [tex]CHM[/tex] and [tex]CHL[/tex]:
[tex]CH=CM\cos 60^{\circ}=\frac{5}{2}, \quad HM=CH\tan 60^{\circ}=\frac{5\sqrt{3}}{2}, \quad HL=CH\tan 30^{\circ}=\frac{5\sqrt{3}}{6}[/tex]
[tex]ML=HM-HL=\frac{5\sqrt{3}}{2}-\frac{5\sqrt{3}}{6}=\frac{5\sqrt{3}}{3}[/tex]

Solution:
The median to the hypotenuse is half of it, so [tex]AB=2CM=\sqrt{17}[/tex]. In the right triangle [tex]CHM[/tex]:
[tex]HM=CM\times \sin \angle HCM=\frac{\sqrt{17}}{2}\times \frac{8}{17}=\frac{4}{\sqrt{17}}[/tex]
So [tex]H[/tex] divides the hypotenuse into the segments
[tex]\frac{\sqrt{17}}{2}+\frac{4}{\sqrt{17}}=\frac{25}{2\sqrt{17}}, \quad\frac{\sqrt{17}}{2}-\frac{4}{\sqrt{17}}=\frac{9}{2\sqrt{17}}[/tex]
The square of each leg equals the hypotenuse times the projection of the leg: [tex]\sqrt{17}\times \frac{9}{2\sqrt{17}}=\frac{9}{2}[/tex] and [tex]\sqrt{17}\times \frac{25}{2\sqrt{17}}=\frac{25}{2}[/tex]. The legs are [tex]\frac{3\sqrt{2}}{2}[/tex] and [tex]\frac{5\sqrt{2}}{2}[/tex].

Solution:
[tex]CC_1=9+16=25[/tex]. The line [tex]AH[/tex] is perpendicular to [tex]BC[/tex], so [tex]\angle HAC_1=90^{\circ}-\angle B[/tex] and [tex]\angle AHC_1=\angle B[/tex]. In the right triangle [tex]AC_1H[/tex]: [tex]\tan B=\frac{AC_1}{16}[/tex]. In the right triangle [tex]AC_1C[/tex]: [tex]\tan A=\frac{25}{AC_1}[/tex]. Then
[tex]\frac{\tan B}{\tan A}=\frac{AC_1^2}{400}=4\;\Rightarrow\;AC_1=40, \quad\tan B=\frac{40}{16}=\frac{5}{2}[/tex]
In the right triangle [tex]BC_1C[/tex]: [tex]BC_1=\frac{CC_1}{\tan B}=10[/tex]. So [tex]AB=50[/tex] and [tex]S=\frac{50\times 25}{2}=625[/tex].

Solution:
Let [tex]D[/tex] be the midpoint of [tex]AB[/tex], [tex]AD=1[/tex], [tex]CD=h[/tex] and [tex]\alpha=\angle BAC[/tex]; then [tex]\tan \alpha=h[/tex]. The circumcentre [tex]O[/tex] and the orthocentre [tex]H[/tex] lie on [tex]CD[/tex]. Since [tex]AH\perp BC[/tex], [tex]\angle HAD=90^{\circ}-\alpha[/tex] and [tex]DH=AD\times \cot \alpha=\frac{1}{h}[/tex]. From [tex]OA=OC[/tex]:
[tex]DO^2+1=(h-DO)^2\iff DO=\frac{h^2-1}{2h}[/tex]
The condition [tex]OD=OH[/tex] means that [tex]O[/tex] is the midpoint of [tex]DH[/tex]:
[tex]\frac{h^2-1}{2h}=\frac{1}{2h}\iff h^2=2[/tex]
Then [tex]AC=\sqrt{1+2}=\sqrt{3}[/tex] and [tex]\sin \alpha=\frac{\sqrt{2}}{\sqrt{3}}=\frac{\sqrt{6}}{3}[/tex].