Introduction to Trigonometry Test

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Introduction to Trigonometry Test - 61

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Weekly Quiz Competition

Question 1
1 / -0

Refer to the above figure and find the length of $$AC$$, if the area of triangle $$ABC$$ is $$15$$.

A
$$2.1$$

B
$$4.1$$

C
$$6.2$$

D
$$8.2$$

E
$$9.6$$

Solution

By sine rule, we get $$\dfrac { AC }{ \sin52 } =\dfrac { CB }{ \sin38 } $$ , which implies $$CB = 0.7813 AC$$
Area of triangle is $$\dfrac {1}{2} \times AC \times CB = 0.39 { AC }^{ 2 } = 15$$
$$\Rightarrow {AC}^{2} = 38.46$$
$$\Rightarrow AC = 6.2$$
Question 2
1 / -0

Calculate the value of $$x$$ if the value of $$\theta = 67^{\circ}$$ in the right angled triangle shown in the above figure.

A
$$2.9$$

B
$$7.6$$

C
$$16.5$$

D
$$17.9$$

E
$$18.2$$

Solution

Given, $$\theta= 67^o$$
$$\cos \theta = \dfrac {7}{x}$$
$$\therefore \theta = 67$$ degrees
$$\therefore \cos(67) = \dfrac {7}{x} = 0.39$$
$$\Rightarrow x = \dfrac {7}{(0.39)} = 17.9$$
Question 3
1 / -0

Find the value of $$\tan (3x) \cot (3x) $$ if it is defined.

A
$$-1$$

B
$$\dfrac {\sqrt {3}}{3}$$

C
$$1$$

D
$$\tan^{3} 3x$$

Solution

The value of $$\tan(3x)\cot(3x) $$
$$=\tan(3x) \times \dfrac {1}{\tan(3x)} = 1$$ ..... (As $$\cot \theta =\dfrac {1}{\tan \theta}$$)
Question 4
1 / -0

The length of side $$DF$$, if the area of triangle $$DEF$$ is $$32\sqrt {3}$$ square units is

A
$$8$$

B
$$8\sqrt {3}$$

C
$$16$$

D
$$16\sqrt {3}$$

Solution

Given : Area of $$\triangle DEF=32\sqrt{3}\ sq. units$$
We know that, Area of right angle triangle $$= \dfrac 12 \times base \times height$$
$$\implies 32\sqrt { 3 } =\dfrac { 1 }{ 2 } DE\times DF$$ ...... $$(1)$$

But, $$\dfrac { DE }{ DF } =\tan { { 60 }^{ o } } $$
$$\implies DE=\sqrt { 3 } DF$$
Substituting in $$(1)$$, we get
$$32\sqrt { 3 } =\dfrac { 1 }{ 2 } (\sqrt { 3 } DF)\times DF$$
$$\implies DF^2=64$$
$$\therefore DF=8$$ units.
Question 5
1 / -0

Simplify $$12\sqrt{3}-(8 \cos x)\left(\displaystyle\frac{3\sqrt 3}{2}\cos x\right)$$.

A
$$\sin^2x$$

B
$$12\sqrt 3\sin^2 x$$

C
$$12\sqrt 3-12\sqrt 3\cos x$$

D
$$12\sqrt 3\cos^2 x$$

Solution

The value of $$12 \sqrt3 - (8\cos x)\left (\dfrac { 3\sqrt { 3 } }{ 2 } \cos x\right)$$
$$= 12 \sqrt3 - 12 \sqrt3 \cos ^{ 2 }{ x } $$
$$ = 12 \sqrt3 (1-\cos ^{ 2 }{ x } ) $$
$$= 12 \sqrt3 \sin ^{ 2 }{ x } $$
Question 6
1 / -0

Simplify $$1 - 2\sin^{2}\theta - 2\cos^{2} \theta $$.

A
$$-2$$

B
$$-1$$

C
$$0$$

D
$$1$$

Solution

We will use $${ \sin }^{ 2 }\theta +{ \cos }^{ 2 }\theta =1$$ to simplify the given expression.

So, we have
$$\begin{aligned}{}1 - 2{\sin ^2}\theta - 2{\cos ^2}\theta & = 1 - 2\left( {{{\sin }^2}\theta + {{\cos }^2}\theta } \right)\\& = 1 - 2\left( 1 \right)\\ &= 1 - 2\\ &= - 1\end{aligned}$$

Hence, option $$B$$ is correct.
Question 7
1 / -0

Find the length of $$AC$$ in the above figure.

A
$$2.79$$

B
$$3.49$$

C
$$3.81$$

D
$$4.05$$

E
$$4.26$$

Solution

As per given figure $$ACB$$ is the right angle triangle and $$AB=5$$ and $$\angle CAB=3x$$ and $$\angle ABC=2x$$
In triangle $$ACB$$, we have
$$2x+3x+90=180$$
$$\Rightarrow 5x=180-90$$
$$\Rightarrow x=18$$
Then $$\angle CAB=54$$ and $$\angle ABC=36$$
$$\sin 36^{0}=\dfrac{AC}{5}$$
$$\Rightarrow AC=5\times \sin36^{0}$$
$$\Rightarrow AC=5\times 0.5581=2.79$$
Question 8
1 / -0

In the figure, $$ABCD$$ is a rectangle. Calculate the area of $$ABCD$$.

A
$$\sqrt {2}$$

B
$$\sqrt {3}$$

C
$$\sqrt {6}$$

D
$$2\sqrt {2}$$

E
$$2\sqrt {3}$$

Solution

The diagonal of a rectangle divides it into two equal right triangles,Triangle $$ABC$$ and Triangle $$ADC$$.

In triangle $$ADC$$,

$$\sin 30^0=\dfrac{CD}{2}$$

$$\Rightarrow \dfrac{1}{2}=\dfrac{CD}{2}$$

$$CD=1$$

Applying Pythagorean theorem, we get

$$AC^2=AD^2+CD^2$$

$$\Rightarrow AD^2=2^2-1^2$$

$$\Rightarrow AD=\sqrt3$$

So, height of triangle $$ACD$$, $$CD=1$$

Length of the base of $$\triangle ACD,$$ $$AD=\sqrt3$$

Area of triangle $$ACD$$,

$$=\dfrac{1}{2}\times$$ base $$\times$$ height

$$=\dfrac{1}{2}\times {\sqrt3} \times1$$

$$=\dfrac{\sqrt3}{2}$$

Similarly,the area of right triangle $$ABC$$ $$=\dfrac{\sqrt3}{2}$$

So, the area of rectangle$$=$$ area of right triangle $$ABC$$ $$+$$ Area of triangle $$ACD=$$ $$\dfrac{\sqrt3}{2}$$$$+\dfrac{\sqrt3}{2}=\sqrt3$$.
Question 9
1 / -0

If $$t = 45^{\circ}$$, what is $$\sec (t) \sin (t) - \mathrm{cosec} (t) \cos (t)$$?

A
$$-2$$

B
$$-1$$

C
$$0$$

D
$$\dfrac {\pi}{2}$$

E
None of the above

Solution

For $$t=45^0$$, the value of $$\sec { (t) } \sin { (t) } -\csc { (t) } \cos { (t) }$$ may be determined as follows:
$$\sec { (t) } \sin { (t) } -\csc { (t) } \cos { (t) } =\sec { (45^{ 0 }) } \sin { (45^{ 0 }) } -\csc { (45^{ 0 }) } \cos { (45^{ 0 }) }$$
$$ =\sqrt { 2 } \times \dfrac { 1 }{ \sqrt { 2 } } -\sqrt { 2 } \times \dfrac { 1 }{ \sqrt { 2 } } $$
$$=1-1$$
$$=0$$
Question 10
1 / -0

Find the value of $$\dfrac {\cos 75^{\circ} . \sin 12^{\circ}. \cos 18^{\circ}}{\sin 15^{\circ}. \cos 78^{\circ} . \sin 72^{\circ}}$$

A
$$2$$

B
$$1$$

C
$$0$$

D
$$-1$$

Solution

Given trigonometric expression is $$\dfrac { \cos { 75° } \sin { 12° } \cos { 18° } }{ \sin { 15°\cos { 78° } \sin { 22° } } }$$

We can write, $$\cos\ 75°$$ as $$\cos (90°-15°), \sin\ 12°$$ as $$\sin (90°-78°)$$ and $$\cos\ 18°$$ as $$\cos (90°-72°)$$

Hence,
$$\dfrac { \cos { 75° } \sin { 12° } \cos { 18° } }{ \sin { 15°\cos { 78° } \sin { 22° } } } =\dfrac { \cos { \left( 90°-15° \right) } \sin { \left( 90°-78° \right) } \cos { \left( 90°-72° \right) } }{ \sin { 15°\cos { 78° } \sin { 22° } } } $$

We know that, $$\cos(90°-\theta)=\sin \theta$$ and $$\sin(90°-\theta)=\cos \theta$$

Hence, we get:

$$\dfrac { \sin { 15°\cos { 78° } \sin { 22° } } }{ \sin { 15°\cos { 78° } \sin { 22° } } } $$

$$=1$$

Hence, the answer is $$\dfrac { \cos { 75° } \sin { 12° } \cos { 18° } }{ \sin { 15°\cos { 78° } \sin { 22° } } }=1$$.