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#### **13.18в)** Да се опрости израза:
$$
C = {1 \over \sin^2\alpha} - {\cot^2\alpha(1+\sin^2\alpha)}
$$
## ***Решение:***
### Имаме, че:
$$
\cot^2\alpha = {\cos^2\alpha \over \sin^2\alpha} \space (1)
$$
### и
$$
1 = {\sin^2\alpha \space + \cos^2\alpha} \space (2)
$$
$$
\Longrightarrow
C = 
{1 - \cos^2\alpha \space-\cos^2\alpha\sin^2\alpha \over \sin^2\alpha }\Longrightarrow (2) \Longrightarrow {\sin^2\alpha \space +\cos^2\not\alpha- \cos^2\not\alpha \space-\cos^2\alpha\sin^2\alpha \over \sin^2\alpha} =$$
$$=
{\sin^2\alpha(1 - \cos^2\alpha) \over \sin^2\alpha}=(2)={\sin^2\alpha}
$$
#### **14.35**
![](https://i.imgur.com/9ySym4k.png)
$$
{S_{ABC} \over S_{ABED}} = {2 \over 7}
$$
### Търси се:
$$
{\tan\alpha + \tan{(90-\alpha)} = \space ?}
$$
### Решение:
###### Ще използваме питагорова теорема:
$$
a^2 + b^2 = c^2 \space (1)
$$
$$
\tan \alpha = {a \over b} \space \space и \space \space
\tan {(90 - \alpha)} = {b \over a} \Longrightarrow {\tan\alpha + \tan{(90-\alpha)}} = {a \over b} + {b \over a} = {a^2 + b^2 \over ab} = (1) = {c^2 \over ab}
$$
$$
S_{ABC} = {ab \over 2}
\space \space \space и \space \space \space
{S_{ABED} = c^2}
\Rightarrow {S_{ABC} \over S_{ABED}} ={{ab \over 2} \over c^2} = {2 \over 7}
\Longrightarrow {ab \over c^2} = {4 \over 7}
\Rightarrow
{c^2 \over ab} = {7 \over 4}
$$

### Пролетни 2016 2.09:

![](https://i.imgur.com/wX0qLWM.png)
Дадено:
$$
\bigtriangleup ABC \space \space \space с \space \sphericalangle ACB = 90 ^\circ ; M - среда\space на\space AB; H - \space пета\space на\space височина\space към\space AB.\space
$$
$$ 
т.N \in AB : MN = NC ; NH = HB + BC
$$
**a)** Намерете $\sphericalangle BAC$
**б)** Нека правата, перпендикулярна на NC от B пресича MC в т. K. Намерете $\sphericalangle KNC$
### Решение:
#### Нека т. Е $\in$ AB : EH = BH $\Rightarrow$ EC = BC $\Rightarrow$ NE = EC, защото NE = NH - EH = ~~HB~~ + BC - ~~EH~~ = BC = EC
#### Нека $\sphericalangle BAC$ = $\alpha$ $\Rightarrow$ $\sphericalangle BMC$ = $2\alpha$ $\Rightarrow$ $\sphericalangle BNC$ =  $2\sphericalangle BMC$ = $4\alpha$, защото от условието имаме, че MN = NC
#### От NE = EC $\Rightarrow$ $\sphericalangle BEC$ = $2\sphericalangle BMC$, но $\sphericalangle BEC$ = $\sphericalangle ABC$ $\Rightarrow$ $\sphericalangle ABC$ + $\sphericalangle BAC$ = $\alpha$ + $8\alpha$ = $9\alpha$ = $90^\circ$ $\Longrightarrow$ $\alpha$ = $10^\circ$
#### Нека т.E $\in$ NC : $\sphericalangle QBC$ = $60^\circ$. Знаем, че $\sphericalangle QCB$ = $60^\circ$ $\Rightarrow$ $\bigtriangleup BQC$ е равностранен $\Rightarrow$ QB = QC **(1)** . Нека видим $\bigtriangleup BQM$ и $\bigtriangleup CQM$:

1) (1)
2) MC = MB $\space$$\stackrel{III\spaceпризнак} \Longrightarrow$ $\bigtriangleup BQM \cong \bigtriangleup CQM$ $\Rightarrow$ $\sphericalangle BMQ$ = $\sphericalangle CMQ$ = $10^\circ$. **(2)**
3) MQ - обща

$\sphericalangle KBC$ = $90^\circ - \sphericalangle QCB = 30^\circ = \sphericalangle QBK \Rightarrow BK \in S_{QC} \Rightarrow \sphericalangle KQC = \sphericalangle KCQ = 20^\circ$
Щом $\sphericalangle KQC = 20^\circ$ и $\sphericalangle BMC = 20^\circ$ $\Rightarrow$ четириъгълникът $MNQK$ е вписан $\Rightarrow$ 
$\Rightarrow$ $\sphericalangle CMK$ = $\sphericalangle KNQ$ = $\stackrel\frown{KQ} \over 2$ = $10^\circ$
$$
\sphericalangle KNC = 10^\circ
$$