Introduction to Trigonometry Test

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Introduction to Trigonometry Test - 35

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Question 1
1 / -0

Find the length of $$AB$$ in cm

A
39.25

B
81.96

C
46.32

D
85.34

Solution

Given, $$BC = 30$$

$$\angle AED = \angle ABC = 90$$
Thus, $$DE = BC = 30$$

Now, In $$\triangle ADE$$
$$\tan 45 = \frac{AE}{DE}$$
$$1 = \frac{AE}{30}$$
$$AE = 30$$

Now, In $$\triangle DBE$$
$$\tan 60 = \frac{BE}{DE}$$
$$BE = 30 \sqrt{3}$$

Now, $$AB = BE + AE$$
$$AB = 30\sqrt{3} + 30$$
$$AB = 81.96$$ cm
Question 2
1 / -0

In the given figure; $$BC= 15$$ cm and $$\displaystyle \sin B= \frac{4}{5}$$.
The measure of $$AC$$ in cm is

A
$$10$$

B
$$15$$

C
$$20$$

D
$$19$$

Solution

In $$\triangle ABC$$, $$BC = 15$$, $$\sin B = \dfrac {4}{5}$$

Now, $$\sin B = \dfrac {P}{H}$$
$$\dfrac {4}{5} = \dfrac {AC}{AB}$$
Then, let $$AC = 4x$$ and $$AB = 5x$$

Now, $$AB^2 = AC^2 + BC^2$$
$$(5x)^2 = (4x)^2 + 15^2$$
$$25x^2 = 16x^2 + 225$$
$$9x^2 = 225$$
$$x = 5$$

Hence, $$AC= 20$$ and $$AB = 25$$
Question 3
1 / -0

If in given figure $$AD= 2$$ cm, then value of $$AB$$ ( in cm ) is

A
2

B
3

C
7

D
1

Solution

Given, $$AD \perp CE$$, $$\angle ACD = 45^{\circ}$$, $$AD =2 $$

Now, In $$\triangle ADC$$
$$\tan \angle ACD = \dfrac{P}{B} = \dfrac{AD}{CD}$$
$$\tan 45 = \dfrac{2}{CD}$$
$$CD = 2=AB$$
$$AB=2$$ cm
Question 4
1 / -0

The length of $$AD$$ (in cm.) is

A
$$10$$

B
$$100$$

C
$$75$$

D
$$50$$

Solution

In $$\triangle ABC$$
$$BC = AC$$
Thus, $$\angle ABC = \angle ACB = x$$ (Isosceles triangle Property)

Now, $$\angle ACD = \angle ABC + \angle ACB$$ (Exterior angle property)
$$60 = x + x$$
$$x = 30^{\circ}$$

Now, $$\sin \angle ABC = \cfrac{P}{H}$$
$$\sin 30 = \cfrac{AD}{AB}$$
$$\cfrac{1}{2} = \cfrac{AD}{100}$$
$$AD = 50$$ cm
Question 5
1 / -0

Find $$PQ$$, if $$\displaystyle AB = 150\ m, \angle P = 30^{\circ}$$ and $$\displaystyle \angle Q = 45^{\circ}$$

A
$$978.8$$

B
$$524.4$$

C
$$279.9$$

D
$$409.8$$

Solution

Given, $$AB = 30$$, $$\angle P = 30^{\circ}$$, $$\angle Q= 45^{\circ}$$

In $$\triangle APB$$,
$$\tan P = \cfrac{P}{B}$$
$$\tan 30 = \cfrac{AB}{PB}$$
$$\cfrac{1}{\sqrt{3}} = \cfrac{150}{PB}$$
$$PB = 150 \sqrt{3}$$

In $$\triangle ABQ$$,
$$\tan Q = \cfrac{P}{B}$$
$$\tan 45 = \cfrac{AB}{BQ}$$
$$1 = \cfrac{150}{BQ}$$
$$BQ = 150 $$

Thus, $$AB = PB +BQ$$
$$AB = 150\sqrt{3} + 150$$
$$AB = 409.8$$
Question 6
1 / -0

In the given figure, $$\displaystyle \angle B = 60^{\circ}, \angle C =30^{\circ}, AB=8\ cm$$ and $$\displaystyle BC=25\ cm$$.
Calculate $$BE$$ in cm.

A
4

B
2

C
5

D
7

Solution

$$\angle B = 60$$, $$\angle C = 30$$, $$AB = 8$$, $$BC = 25$$

Now, In $$\triangle ABE$$
$$\cos B = \frac{B}{H} = \frac{BE}{AB}$$
$$\cos 45 = \frac{BE}{8}$$
$$BE = 4$$
Question 7
1 / -0

Find :
$$AD$$

A
2

B
6

C
4

D
1

Solution

Given: $$AB = 12$$, $$\angle ACB = 30^{\circ}$$
$$\angle ACB + \angle ABC + \angle BAC = 180$$
$$30 + 90 + \angle BAC = 180$$
$$\angle BAC = 60^{\circ}$$

Now, In $$\triangle ABD$$
Now, $$\cos 30 = \frac{B}{H} = \frac{AD}{AB}$$
$$\frac{1}{2} = \frac{AD}{12}$$
$$AD = 6$$
Question 8
1 / -0

$$(1+tan \alpha \ tan\beta)^2+(tan\alpha -tan\beta)^2$$ is equal to

A
$$tan^2\alpha +tan^2\beta$$

B
$$cos^2\alpha cos^2\beta$$

C
$$sec^2\alpha sec^2\beta$$

D
$$tan^2\alpha tan^2\beta$$

Solution

$$(1+tan \alpha tan\beta)^2+(tan\alpha -tan\beta)^2$$

$$=1+tan^2\alpha tan^2\beta+tan^2\alpha+tan^2\beta+2tan \alpha tan\beta-2tan \alpha tan\beta$$

$$=1+tan^2\alpha tan^2\beta+tan^2\alpha+tan^2\beta$$

$$=tan^2\alpha (1+tan^2\beta)+1(1+tan^2\beta)$$

$$=(1+tan^2\alpha)(1+tan^2\beta)$$

$$=sec^2\alpha sec^2\beta$$
Question 9
1 / -0

$$6(sin^6\theta+cos^6\theta)-9(sin^4\theta+cos^4\theta)$$ is equal to

A
$$-1$$

B
$$1$$

C
$$-3$$

D
$$3$$

Solution

Consider $$6(sin^6\theta+cos^6\theta)-9(sin^4\theta+cos^4\theta)$$

$$=6[(sin^{ 2 }\theta )^{ 3 }+(cos^{ 2 }\theta )^{ 3 }]-9[(sin^{ 2 }\theta )^{ 2 }+(cos^{ 2 }\theta )^{ 2 }]$$

$$=6[(sin^{ 2 }\theta +cos^{ 2 }\theta )^{ 3 }-3sin^{ 2 }\theta cos^{ 2 }\theta (sin^{ 2 }\theta +cos^{ 2 }\theta )]-9[(sin^{ 2 }\theta +cos^{ 2 }\theta )^{ 2 }-2sin^{ 2 }\theta cos^{ 2 }\theta ]$$

$$=6(1-3sin^{ 2 }\theta cos^{ 2 }\theta )-9(1-2sin^{ 2 }\theta cos^{ 2 }\theta )$$

$$=6-9=-3$$
Question 10
1 / -0

If $$\sin x+\sin^2x+\sin^3x=1$$, then $$\cos^6x-4\cos^4x+8 \cos^2x$$ is equal to

A
$$0$$

B
$$2$$

C
$$4$$

D
$$8$$

Solution

$$\sin x+{ \sin }^{ 2 }x+{ \sin }^{ 3 }x=1\Rightarrow \sin x+{ \sin }^{ 3 }x=1-{ \sin }^{ 2 }x\\ \Rightarrow \sin x\left( 1+\sin ^{ 2 }x \right) ={ \cos }^{ 2 }x\Rightarrow \sin x\left( 2-{ \cos }^{ 2 }x \right) ={ \cos }^{ 2 }x\\ \Rightarrow { \sin }^{ 2 }x{ \left( 2-{ \cos }^{ 2 }x \right) }^{ 2 }={ \cos }^{ 4 }x\\ \Rightarrow \left( 1-{ \cos }^{ 2 }x \right) \left( 4+{ \cos }^{ 4 }x-4{ \cos }^{ 2 }x \right) ={ \cos }^{ 4 }x\\ \Rightarrow 4+{ \cos }^{ 4 }x-4{ \cos }^{ 2 }x-4{ \cos }^{ 2 }x-{ \cos }^{ 6 }x+4{ \cos }^{ 4 }x={ \cos }^{ 4 }x\\ \Rightarrow { \cos }^{ 6 }x-4{ \cos }^{ 4 }x+8{ \cos }^{ 2 }x=4$$