Introduction to Trigonometry Test

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Introduction to Trigonometry Test - 37

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Weekly Quiz Competition

Question 1
1 / -0

If $$\sin C$$ is $$\displaystyle \frac{1}{m}$$, then $$m$$ is:

A
$$1$$

B
$$5$$

C
$$2$$

D
$$3$$

Solution

Given, $$\triangle ABC$$, A perpendicular from B on AC.
Let it cut AC at D, such that$$\angle BDA = \angle BDC = 90^{\circ}$$
$$AD = 3$$
$$BD = 4$$
$$BC = 12$$

In $$\triangle ABD$$,
$$AB^2 = BD^2 + AD^2$$
$$AB^2 = 3^2 + 4^2$$
$$AB = 5$$

In $$\triangle BCD$$
$$BC^2 = CD^2 + BD^2$$
$$12^2 = CD^2 + 4^2$$
$$CD^2 = 128$$
$$CD = 8 \sqrt{2}$$

Now, $$\sin C = \dfrac{P}{H} = \dfrac{BD}{DC} = \dfrac{4}{12} = \dfrac{1}{3}$$

So, option D is correct.
Question 2
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If $$\cos A$$ is $$\displaystyle \frac{3}{m}$$, then m is:

A
$$5$$

B
$$1$$

C
$$2$$

D
$$7$$

Solution

Given, $$\triangle ABC$$, A perpendicular from B on AC, let it cut AC at D, such that$$\angle BDA = \angle BDC = 90^{\circ}$$
$$AD = 3$$
$$BD = 4$$
$$BC = 12$$

In $$\triangle ABD$$,
$$AB^2 = BD^2 + AD^2$$
$$AB^2 = 3^2 + 4^2$$
$$AB = 5$$

In $$\triangle BCD$$
$$BC^2 = CD^2 + BD^2$$
$$12^2 = CD^2 + 4^2$$
$$CD^2 = 128$$
$$CD = 8 \sqrt{2}$$

Now, $$\cos A = \dfrac{B}{H} = \dfrac{AD}{AB} = \dfrac{3}{5}$$
$$\Longrightarrow$$ $$m=5$$
Question 3
1 / -0

If $$cos x+sin x=\sqrt 2 cos x$$, then $$tan^2x+2 tan x$$ is equal to

A
$$0$$

B
$$1$$

C
$$2$$

D
$$3$$

Solution

Given $$cos x+sin x=\sqrt 2 cos x$$

$$(cos x+sin x)^2=2 cos^2 x$$
$$cos^2x+sin^2x+2sinxcosx=2cos^2x$$
Divide $$cos^2x$$ both sides
$$\Rightarrow 1+\tan^2x+2 \tan x=2$$

$$\Rightarrow \tan^2x+2 \tan x=1$$
Question 4
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Given: $$\cos A\, =\, \displaystyle \cfrac{5}{13}$$. If $$\displaystyle \frac{\sin A\, -\, \cot A}{2\, \tan\, A}$$ is $$\displaystyle \frac{m}{3744}$$, $$m$$ is:

A
695

B
595

C
295

D
395

Solution

Given, $$\cos A = \frac{5}{13}$$
$$\cos A = \frac{B}{H} = \frac{5}{13}$$

Now, using Pythagoras Theorem,
$$H^2 = P^2 + B^2$$
$$13^2 = P^2 + 5^2$$
$$P = 12$$

Now, $$\displaystyle \frac{\sin A\, -\, \cot A}{2\tan A}$$

= $$\displaystyle \frac{\frac{12}{13}\, -\, \frac{5}{12}}{2 (\frac{12}{5})}$$

= $$\displaystyle \frac{\frac{144 - 65}{156}}{\frac{24}{5}}$$

= $$\displaystyle \frac{\frac{79}{156}}{\frac{24}{5}}$$

= $$\displaystyle \frac{395}{3744}$$
Question 5
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If $$ 5 \cot\, \theta\, =\, 12$$, find the value of : $$\csc\, \theta\, +\, \sec \, \theta$$ is $$\displaystyle {3}\cfrac{41}{m}$$, m is

A
$$40$$

B
$$60$$

C
$$50$$

D
$$13$$

Solution

$$5 \cot \theta = 12$$
$$\cot \theta = \cfrac{12}{5}$$
$$\cot \theta = \cfrac{B}{P} = \cfrac{12}{5}$$

Using Pythagoras Theorem,
$$H^2 = P^2 + B^2$$
$$H^2 = 12^2 + 5^2$$
$$H = 13$$

Now, $$\csc \theta + \sec \theta$$
= $$\cfrac{H}{P} + \cfrac{H}{B}$$
= $$\cfrac{13}{12} + \cfrac{13}{5}$$
= $$\cfrac{65 + 156}{60}$$
= $$\cfrac{221}{60}$$
= $$3 \cfrac{41}{60}$$
Question 6
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Directions For Questions

In the given figure, triangle ABC is right-angled at B. D is the foot of the perpendicular from B to AC. Given that BC$$=$$3 cm and AB$$=$$4 cm. Without using tables, find :

...view full instructions

$$sin\, \angle DBA$$ is $$\displaystyle \frac{4}{m}$$
value of m is

A
0

B
7

C
9

D
5

Solution

In $$\triangle ABC$$
$$AB^2 + BC^2 = AC^2$$ (By Pythagoras theorem)
$$4^2+ 3^2 = AC^2$$
$$AC^2 = 25$$
$$AC =5$$

In $$\triangle ABC$$
$$\angle A + \angle B + \angle C = 180$$
$$\angle A + \angle C = 90$$ (As $$\angle B = 90$$).....(i)

In $$\triangle ABD$$,
$$\angle A + \angle BDA + \angle DBA = 180$$
$$\angle A + \angle DBA = 90$$ (As $$\angle BDA = 90$$).....(i)

From (i) and (ii)
$$\angle C = \angle DBA$$
Similarly, $$\angle DBC = \angle A$$

Now, $$\sin \angle DBA = \sin \angle C = \frac{P}{H} = \frac{AB}{AC}= \frac{4}{5}$$
Question 7
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Directions For Questions

In the given figure, triangle ABC is right-angled at B. D is the foot of the perpendicular from B to AC. Given that BC$$=$$3 cm and AB$$=$$4 cm. Without using tables, find :

...view full instructions

$$tan\, \angle DBC$$ is $$\displaystyle \frac{m}{4}$$
value of m is

A
1

B
3

C
4

D
2

Solution

In $$\triangle ABC$$
$$AB^2 + BC^2 = AC^2$$ (By Pythagoras theorem)
$$4^2+ 3^2 = AC^2$$
$$AC^2 = 25$$
$$AC =5$$

In $$\triangle ABC$$
$$\angle A + \angle B + \angle C = 180$$
$$\angle A + \angle C = 90$$ (As $$\angle B = 90$$).....(i)

In $$\triangle ABD$$,
$$\angle A + \angle BDA + \angle DBA = 180$$
$$\angle A + \angle DBA = 90$$ (As $$\angle BDA = 90$$).....(i)

From (i) and (ii)
$$\angle C = \angle DBA$$
Similarly, $$\angle DBC = \angle A$$

Now, $$\tan \angle DBC = \tan \angle A = \dfrac{P}{B} = \dfrac{BC}{AB}= \dfrac{3}{4}$$
Question 8
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Given : $$sec\, A\, =\, \displaystyle \frac{29}{21}$$, evaluate : $$sin\, A\, -\, \displaystyle \frac{1}{tan\, A}$$ is $$\displaystyle -\frac{m}{580}$$, m is

A
130

B
215

C
209

D
524

Solution

$$\sec A = \frac{29}{21}$$
$$\sec A = \frac{H}{B} = \frac{29}{21}$$

Using Pythagoras Theorem,
$$H^2 = P^2 + B^2$$
$$29^2 = P^2 + 21^2$$
$$841 = 441 + P^2$$$
$$P = 20$$

Now, $$\sin A - \frac{1}{\tan A} = \sin A - \cot A$$
= $$\frac{P}{H} - \frac{B}{P}$$
= $$\frac{20}{29} - \frac{21}{20}$$
= $$\frac{400 - 609}{580}$$
= $$\frac{- 209}{580}$$
Question 9
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In $$\triangle ABC$$ If $$\sin B$$ is $$\displaystyle \frac{12}{m}$$, then $$m$$ is:

A
13

B
12

C
11

D
10

Solution

In $$\triangle ABC$$, $$AD \perp BC$$ $$AB = 13$$, $$BD = 5$$, $$DC =16$$

Now, In $$\triangle ABD$$,
$$AB^2 = AD^2 + BD^2$$
$$13^2 = AD^2 + 5^2$$
$$AD^2 = 144$$
$$AD = 12$$

Now, in $$\triangle ADC$$,
$$AC^2 = AD^2 + CD^2$$
$$AC^2 = 12^2 + 16^2$$
$$AC^2 = 144 + 256$$
$$AC = 20$$ cm

Now, $$\sin B = \frac{P}{H} = \frac{AD}{AB} = \frac{12}{13}$$
Question 10
1 / -0

If $$\tan C$$ is $$\displaystyle \frac{3}{m}$$, then $$m$$ is:

A
1

B
4

C
3

D
2

Solution

In $$\triangle ABC$$, $$AD \perp BC$$ $$AB = 13$$, $$BD = 5$$, $$DC =16$$

Now, In $$\triangle ABD$$,
$$AB^2 = AD^2 + BD^2$$
$$13^2 = AD^2 + 5^2$$
$$AD^2 = 144$$
$$AD = 12$$

Now, in $$\triangle ADC$$,
$$AC^2 = AD^2 + CD^2$$
$$AC^2 = 12^2 + 16^2$$
$$AC^2 = 144 + 256$$
$$AC = 20$$ cm

Now, $$\tan C = \frac{P}{B} = \frac{AD}{CD} = \frac{12}{16} = \frac{3}{4}$$