# Quiz. Trigonometry.

1-In the statements below find which is True and which is False

1. sin1790<0
2. cos1780<0
3. sin450=cos450
4. sin70=cos830
5. sinα=1,1
6. sin2 0+cos2 0=0
7. tg1010 <0
8. cotg1300 >0

2-Given $S=(sinx+cosx)^{2}-(sinx-cosx)^{2}$ , verify that the area   $S=4sinx\cdot&space;cosx$ and find the value of S when  x=300 .

3-The angles of a triangle are  45dhe 75. Find the measure of the third angle in radian .

Find the third angle in degrees 1800-(450+750)=600.Then plug in  $\frac{\alpha&space;}{\pi&space;}=\frac{a}{180^{0}}$ . The third triangle is $\frac{\pi&space;}{3}$

4-Find the domain of the function   $y=\sqrt{cosx}$ .  Answer: $x\epsilon&space;[0^{0};90^{0}]$

5-Verify the indentity.$tg\alpha&space;+cotg\alpha&space;=\frac{1}{sin\alpha&space;\cdot&space;cos\alpha&space;}$ .

Answer $tg\alpha&space;+cotg\alpha&space;=\frac{sin\alpha&space;}{cos\alpha&space;}+\frac{cos\alpha&space;}{sin\alpha&space;}=\frac{sin^{2}\alpha&space;+cos^{2}\alpha&space;}{sin\alpha&space;\cdot&space;cos\alpha&space;}=\frac{1}{sin\alpha&space;\cdot&space;cos\alpha&space;}$

6- Given  b=2; $c=\sqrt{2}$; dhe β=450  in the picture, find the angles γ and α.

Based on the sin theorem we can write $\frac{c}{sin\gamma&space;}=\frac{b}{sin\beta&space;}\Rightarrow&space;sin\gamma&space;=\frac{c\cdot&space;sin\beta&space;}{b}=\frac{\sqrt{2}\cdot&space;\frac{\sqrt{2}}{2}}{2}=\frac{1}{2}\Rightarrow&space;\gamma&space;=30^{0}$

$\alpha&space;=180^{0}-(\beta&space;+\gamma&space;)=.....$

7-Given triangle with  a=5cm; b=6cm dhe c=7cm find the Area(S) of the triangle..

Find the radius of the circumcenter of the tringle. Find the sin and cos of the angle α.

$p=\frac{a+b+c}{2}=\frac{5+6+7}{2}=9$ . We can find the area by using Heron Formula $S=\sqrt{p(p-a)(p-b)(p-c)}=\sqrt{9(9-5)(9-6)(9-7)}=\sqrt{9\cdot&space;4\cdot&space;3\cdot&space;2}=6\sqrt{6}$ .

From the sin theorem we write: $\frac{a}{sin\alpha&space;}=2R\Rightarrow&space;sin\alpha&space;=\frac{a}{2R}=\frac{5}{2\cdot&space;\frac{35\sqrt{6}}{6}}=\frac{3}{7\sqrt{6}}=\frac{3\sqrt{6}}{42}=\frac{\sqrt{6}}{14}$ . From the cos theorem we write : $cos\alpha&space;=\sqrt{1-sin^{2}\alpha&space;}=......$ .We can see that the angle is an acute angle.

8-In the triangle  ABC there are givenb=4; $c=\sqrt{2}$; α=450.Find: the side  a; sinβ ; sinγ; the height of the triangle taken from the C apex .

Using the cos theorem we can find the side a. $a^{2}=b^{2}+c^{2}-2bc\cdot&space;cos\alpha&space;=4+2-4\sqrt{2}\cdot&space;\frac{\sqrt{2}}{2}=2$ . Using sin theorem we can find sinβ dhe sinγ.

To find the height of the triangle taken from the  C apex , first we can find the area of the triangle S. $S=\frac{1}{2}bc\cdot&space;sin\alpha&space;=\frac{1}{2}4\cdot&space;\sqrt{2}\cdot&space;\frac{\sqrt{2}}{2}=2.$ We also know that  $S=\frac{1}{2}c\cdot&space;h\Rightarrow&space;h=\frac{2S}{c}=\frac{2\cdot&space;2}{\sqrt{2}}=2\sqrt{2}$. where h is the height of the triangle taken from the C apex.