Quiz. Trigonometry.

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– Posted on 29/03/2014Posted in: MATH

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 cosx and find the value of S when  x=300 .

3-The angles of a triangle are  450 dhe 750 . Find the measure of the third angle in radian .

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

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

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

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

zgjidhja e trekendeshit16- 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 }=\frac{b}{sin\beta }\Rightarrow sin\gamma =\frac{c\cdot sin\beta }{b}=\frac{\sqrt{2}\cdot \frac{\sqrt{2}}{2}}{2}=\frac{1}{2}\Rightarrow \gamma =30^{0}

\alpha =180^{0}-(\beta +\gamma )=.....

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 4\cdot 3\cdot 2}=6\sqrt{6} .

From the sin theorem we write: \frac{a}{sin\alpha }=2R\Rightarrow sin\alpha =\frac{a}{2R}=\frac{5}{2\cdot \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 =\sqrt{1-sin^{2}\alpha }=...... .We can see that the angle is an acute angle.

zgjidhja e trekendeshit38-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 cos\alpha =4+2-4\sqrt{2}\cdot \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 sin\alpha =\frac{1}{2}4\cdot \sqrt{2}\cdot \frac{\sqrt{2}}{2}=2. We also know that  S=\frac{1}{2}c\cdot h\Rightarrow h=\frac{2S}{c}=\frac{2\cdot 2}{\sqrt{2}}=2\sqrt{2}. where h is the height of the triangle taken from the C apex.

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