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Trigonometric Ratios and Height and Dis...

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Trigonometric Ratios and Height and Distance Practice Quiz

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“Trigonometry (from Greek trigōnon, “triangle” and metron, “measure”) is a branch of mathematics that studies relationships between side lengths and angles of triangles.”Trigonometry has found its importance in multiple fields. It has found its applications in the fields ranging from Engineering to Architecture to Astronomy as well. Trigonometry can be used in these fields by to measure distances and angles by assuming lines that connect the points. As a result, Trigonometric Ratios and Height and Distance are interconnected.


The article Trigonometric Ratios and Height and Distance Practice Quiz lists different types of Trigonometric Ratios and Height and Distance questions with solutions useful for candidates preparing for different competitive examinations like RRB ALP/Technical Exams/Junior Engineer Recruitment Exams, SSC, CAT and etc. [What is a Triangle?]



shape Quiz

1. If sin θ + cosec θ = 2, then \({sin}^{2}θ + {cosec}^{2}θ \)is equal to:

    A. 1
    B. 4
    C. 2
    D. None of these


Answer – Option C

Explanation –

Given that, sinθ + cosecθ = 2

On squaring both sides, we get

\({sin}^{2}θ + {cosec}^{2}θ \) + 2 = 4

\({sin}^{2}θ + {cosec}^{2}θ \) = 2

2. The value of \(\frac{{cot}^{2}θ + 1}{{cot}^{2}θ – 1}\) is equal to:

    A. sin 2θ
    B. cos 2θ
    C. cosec 2θ
    D. sec 2θ


Answer – Option D

Explanation –

\(\frac{{cot}^{2}θ + 1}{{cot}^{2}θ – 1}\) = \(\frac{+ 1{tan}^{2}θ}{1 – + 1{tan}^{2}θ}\)

\(\frac {1}{{cos}^{2}θ + {sin}^{2}θ} \) = \(\frac{1}{cos2θ}\) = sec2θ

3. If tan A = 2 tan B + cot B, then 2 tan (A – B) is equal to:

    A. tan B
    B. 2 tan B
    C. cot B
    D. 2 cot B


Answer – Option C

Explanation –

Given that

tan A = 2 tan B + cot B …(i)

Now, 2 tan (A – B) = 2(\(\frac{tan A – tan B}{1 + tanA tan B}\))

\(2 \frac{(2tan B + cot B – tan B)}{1 + (2tan B cot B) tan B}\)

\(2 \frac{tan B cot B}{2(1 + {tan}^{2} B)}\) = \(\frac{cotB({tan}^{2} B + 1)}{1 + {tan}^{2} B}\) = cotB

4. If tan A – tan B = x and cot B – cot A = y, then cot (A – B) is equal to:

    A. \(\frac{1}{x} + y\)
    B. \(\frac{1}{xy}\)
    C. \(\frac{1}{x} – \frac{1}{y}\)
    D. \(\frac{1}{x} + \frac{1}{y}\)


Answer – Option D

Explanation –

Given that

tan A – tan B = x …(i)

and cot B – cot A = y …(ii)

Now, cot (A – B) = \(\frac{1}{tan (A – B)}\)

= \(\frac{1 + tan A tan B}{tan A – tan B}\)

= \(\frac{1}{tan A – tan B)} + \frac{tan A tan B}{tan A – tan B}\)

= \(\frac{1}{x} + \frac{1}{y}\) [from (i) and (ii)]

5. If sin θ = \(\frac{4}{5}\) and θ lies in the third quadrant, then cos \(\frac{θ}{2}\) is equal to:

    A. \(\frac{1}{\sqrt{5}}\)
    B. \( – \frac{1}{\sqrt{5}}\)
    C. \(\sqrt{\frac{2}{5}}\)
    D. –\(\sqrt{\frac{2}{5}}\)


Answer – Option B

Explanation –

Given that,

sinθ = \( – \frac{4}{5}\) = and θ lies in the IIIrd quadrent

cosθ = \(\sqrt{1 – \frac{16}{25}}\)

Now, \( cos \frac{θ}{2}\) = \(\sqrt{ \frac{1 + cosθ}{2}}\) = \(\sqrt{ \frac{1 – \frac{3}{5} }{2}}\) = ± \(\sqrt{\frac{1}{5}}\)

But we take \( cos \frac{θ}{2}\) = –\(\sqrt{\frac{1}{5}}\) Since, if θ lies in IIIrd

quadrant, then \(\frac{θ}{2}\) will be in IInd quadrant.

Hence, \( cos \frac{θ}{2}\) = –\(\sqrt{\frac{1}{5}}\)

6. The value of cos 1º cos 2º cos 3º… cos 100º is equal to

    A. 0
    B. -1
    C. 0
    D. None of these


Answer – Option C

Explanation –

cos 1º cos 2º cos 3º … cos 90º… cos 100º

= cos 1º cos 2º cos 3º … 0 … cos 100º = 0

[i.e, cos 90º = 0]

7. The value of sin 12º sin 48º sin 54º is equal to:

    A. \(\frac{1}{16}\)
    B. \(\frac{1}{32}\)
    C. \(\frac{1}{8}\)
    D. \(\frac{1}{4}\)


Answer – Option C

Explanation –

Now, sin 12º sin 48º sin 54º

\(\frac{1}{2}\)(cos36º – cos60º )cos36º

\(\frac{1}{2}(\frac{(\sqrt{5} + 1}{4}\frac{1}{2})(\frac{(\sqrt{5} + 1}{4})\) = \(\frac{5 -1}{32} = \frac{4}{32} = \frac{1}{8}\)

8. The value of 2 cos x – cos 3x – cos 5x is equal to:

    A. \(16{cos}^{3}x {sin}^{2}x\)
    B. \(16{sin}^{3}x {cos}^{2}x\)
    C. \(4{cos}^{3}x {sin}^{2}x\)
    D. \(4{sin}^{3}x {cos}^{2}x\)


Answer – Option A

Explanation –

2 cos x – cos 3x – cos 5x = 2 cos x – 2 cos x cos 4x

= 2 cos x (1 – cos 4x) = 2 cos x 2 \({sin}^{2}2x\)

4 cos x\({ (2 sin x cos x)}^{2}\) = \(16 {sin}^{2}x{cos}^{3}x\)

9. If cos θ = \(\frac{1}{2}(x + \frac{1}{x})\) then \(\frac{1}{2}({x}^{2} + \frac{1}{{x}^{2}})\) is equal to:

    A. sin 2θ
    B. cos 2θ
    C. tan 2θ
    D. sec 2θ


Answer – Option D

Explanation –

Given that cos θ = \(\frac{1}{2}(x + \frac{1}{x})\) = \(x + \frac{1}{x}\) = 2 cosθ

We know that \({x}^{2} + \frac {1}{{x}^{2}}\) = \({x}^{2} + \frac {1}{{x}^{2}}\)

= \({2cos θ}^{2} -2 = {4 cos θ}^{2}\) -2

= 2cos 2θ [from (i)]

i.e, \(\frac{1}{2}({x}^{2} + \frac {1}{{x}^{2}})\) =\( \frac{1}{2} * {x}^{2} 2cos 2θ\)

10. The value of x for the maximum value of \(\sqrt{3}\)cosx + sin x is:

    A. 30º
    B. 45º
    C. 60º
    D. 90º


Answer – Option A

Explanation –

Let f(x) = \(\sqrt{3}\)cosx + sin x

f(x) = \(2(\frac{\sqrt{3}}{2}cosx + \frac{1}{2}sinx) \) = 2sin (\( x + \frac{\pi}{3}\))

Since, -1 ≤ sin (x + \(\frac{\pi}{3}) ≤ 1\)

Hence, f(x) is maximum if x + \(\frac{\pi}{3}\) = \(\frac{\pi}{2}\)

x = \(\frac{\pi}{6}\) = 30º

11. The equation \({(a + b)}^{2}\) = 4ab \({sin}^{2}θ \) is possible only when

    A. 2a = b
    B. a = b
    C. a = 2b
    D. None of these


Answer – Option A

Explanation –

We have \({(a + b)}^{2}\) = 4ab \({sin}^{2}θ \)

\({sin}^{2}θ \) = \(\frac{{(a + b)}^{2}}{4ab}\)

Since \({sin}^{2}θ \) ≤ 1

= \(\frac{{(a + b)}^{2}}{4ab}\) ≤ 1

= \({(a + b)}^{2} – 4ab = {(a + b)}^{2} \) ≤ 1

= \({(a – b)}^{2} \) ≤ 0

a = b

12. The value of the expression 1 – \(\frac{{sin}^{2}y}{1 + cos y} + \frac{1 + cos y}{sin y} – \frac {sin}{1 – cos y}is equal to:\)

    A. 0
    B. 1
    C. sin y
    D. cos y


Answer – Option D

Explanation –

The given expression can be written as

= \(\frac { 1 + cos y – {sin} ^ {2}y }{1 + cos y}\) + \( \frac { 1 + {cos}^{2} y – {sin}^{2} y }{sin y (1 + cos y)} \)

= \(\frac{cos y (1 + cos y)}{1 + cos y} + 0\) = cos y

13. The circular wire of diameter 10 cm is cut and placed along the circumference of a circle of diameter 1m. The angle subtended by the write at the center of the circle is equal to:

    A. \(\frac{\pi}{4}rad\)
    B. \(\frac{\pi}{3}rad\)
    C. \(\frac{\pi}{5}rad\)
    D. \(\frac{\pi}{10}rad\)


Answer – Option C

Explanation –

Given that, diameter of circular wire = 10 cm,

Length of wire = 10\(\ pi\)

Hence, required angle = \(\frac{length of arc}{radius of big circle}\)

\(\frac{10 \pi}{50}\) = \(\frac{\pi}{5}\)

14. The greatest and least value of sin x cos x are:

    A. 1, -1
    B. \(\frac{1}{2}\), – \(\frac{1}{2}\)
    C. \(\frac{1}{4}\), – \(\frac{1}{4}\)
    D. 2, -2


Answer – Option B

Explanation –

f(x) = sin x cos x = \(\frac{1}{2} sin 2x\)

We know – 1 ≤ sin 2x ≤ 1

– \(\frac{1}{2}\) ≤ \(\frac{1}{2} sin 2x\) ≤ \(\frac{1}{2}\)

Thus, the greatest and least value of f(x) are \(\frac{1}{2}\) and –\(\frac{1}{2}\) respectively

15. If A = \({sin}^{2}θ + {cosec}^{4}θ \), then for all real values of θ

    A. 1 ≤ A ≤ 2
    B. \(\frac{3}{4}\) ≤ A ≤ 1
    C. \(\frac{13}{16}\) ≤ A ≤ 1
    D. \(\frac{3}{4}\) ≤ A ≤ \(\frac{13}{16}\)


Answer – Option B

Explanation –

We have, A = \({sin}^{2}θ + {cos}^{4}θ \)

= \({sin}^{2}θ + {cos}^{2}θ {cos}^{2}θ \) ≤ \({sin}^{2}θ + {cos}^{2}θ \) (Since, \({cos}^{2}θ\) ≤ 1)

\({sin}^{2}θ + {cos}^{4}θ \) ≤ 1

A ≤ 1

Again, \({sin}^{2}θ + {cos}^{4}θ \) = 1 – \({cos}^{2}θ + {cos}^{4}θ \)

= \({cos}^{4}θ – {cos}^{2}θ + 1 \)

= \({{cos}^{2}θ – \frac{1}{2}}^{2}\) + \(\frac{3}{4}\) ≥ \(\frac{3}{4}\)

Hence \(\frac{3}{4}\)≤ A ≤ 1

1. The value of \( {sin}^{2} 30º + {sin}^{2} 60º\) is

    A. 1
    B. \(\frac{3}{2}\)
    C. 2
    D. \(\frac{3}{4}\)


Answer – Option A

Explanation –

\( {sin}^{2} 30º + {sin}^{2} 60º\) = \( {sin}^{2} 30º + {cos}^{2} 30º\) = 1

2. tan 90° is undefined. As θ is increased from 89° towards 90°, value of tan θ tends to

    A. 0
    B. +oo
    C. 1
    D. undefined


Answer – Option B

Explanation –

Value of tan q will ten A to + ∞ as tan 90° is + ∞

3. The angle of elevation of a ladder leaning against a wall is 60° i.e. ladder makes an angle of 60° with the ground. The foot of the ladder is 4.6 metres away from the wall. What is the length of this ladder?

    A. 9.2 m
    B. 2.3 m
    C. 6.9 m
    D. 7.8 m


Answer – Option A

Explanation –

\(\frac{4.2}{L}\) = COS 60º

L = 4.6 × 2 = 9.2 m

4. A kite is flying at a height of 75 m from the level ground, attached to a string inclined at 60° to the horizontal. The length of the string is

    A. 50 \(\sqrt{2}\) m
    B. 50 \(\sqrt{3}\) m
    C. \(\frac{50}{\sqrt{2}}\)
    D. \(\frac{50}{\sqrt{3}}\)


Answer – Option B

Explanation –

\(\frac{75}{L}\) = cos60º

L = \(\frac{75 * 2}{\sqrt{3}}\) = 50 \(\sqrt{3}\) m

5. A tower stands vertically on the ground. From a point on the ground which is 30 m away from the foot of the lower, the angle of elevation of the top of the tower is 60°. The height of the tower is:

    A. 30 \(\sqrt{3}\) m
    B. 10 \(\sqrt{3}\) m
    C. 15m
    D. 6m


Answer – Option A

Explanation –

Figure

tan 60° = \(\frac{Height}{30}\)

Height of the tower = 30 \(\sqrt{3}\) m

6. From a point on the bridge across a river the angles of depression of the banks on the opposite side of the river are 30° and 45° respectively. If the bridge is at a height of 4 m from the bank, the width of the river is

    A. 4 (\(\sqrt{3} – 1\)) m
    B. 2 (\( 2\sqrt{3} – 1\)) m
    C. 4 (\(\sqrt{3} + 1\)) m
    D. 2 (\( 2\sqrt{3} + 1\)) m


Answer – Option C

Explanation –

Figure

In triangle ABD, tan 30° = \(\frac{DB}{AB}\)

AB = 4 \(\sqrt{3}\) m

Similarly, in triangle CBD, tan 45° = \(\frac{DB}{BC}\)

BC = DB = 4 meter

Hence, required distance AC = 4 \(\sqrt{3}\) + 4

= 4 (\(\sqrt{3} + 1\)) m

7. A ladder 15 m long just reaches the top of a vertical wall. If the ladder makes an angle of 60°
with the wall, the height of the wall is


    A. 7.5 m
    B. 5 \( \sqrt{3}\)) m
    C. 15 (\(\frac {\sqrt{3}}{2}\)) m
    D. 10 (\( \sqrt{3}\)) m


Answer – Option A

Explanation –

cos60 = \(\frac{h}{15}\)

h = 15 * \(\frac{1}{2}\) = 7.5 m

8. From a point on the bridge across a river the angles of depression of the banks on the opposite
side of the river are 30° and 45° respectively. If the bridge is at a height of 3m from the bank, the width of the river is


    A. 2 (\( \sqrt{3} + 1\)) m
    B. 3 (\( \sqrt{3} + 1\)) m
    C. 4(\( \sqrt{3} + 1\)) m
    D. 2 (\( 2\sqrt{3} + 1\)) m


Answer – Option B

Explanation –


According to the figure.

In △ACD, \(\frac{AC}{CD}\) = tan 45º

AC = CD

or CD = 3m

Again in △ACB, \(\frac{AC}{BC}\) = tan 30º

BC = 3(\( \sqrt{3}\))

Hence, BD = 3 (\( \sqrt{3}\)) + 3(\( \sqrt{3} + 1\)) m

9. A ladder just reaches the top of a wall. The foot of the ladder is 8 m away from the foot of the wall. The ladder makes an angle of 60° with the ground. The length of the ladder is

    A. 4 m
    B. 16 m
    C. 16 \(\frac{\sqrt{3}}{3} \)
    D. 16 \( \sqrt{3} \)


Answer – Option B

Explanation –


cos 60° = \(\frac{8}{1}\)

\(\frac{1}{2}\) = \(\frac{8}{2}\)

1 = 16 m

10. From a point on the bridge across a river the angle of depressions of the bank on the opposite
side of the river are 60° and 45° respectively. If the bridge is at a height of 3 m from the bank, the width of the river is


    A. 2 (\( \sqrt{3} + 1\)) m
    B. 3 (\( \sqrt{3} + 1\)) m
    C. 3 – (\( \sqrt{3} \)) m
    D. 3 + (\( \sqrt{3} \)) m


Answer – Option D

Explanation –


From the figure, tan 60° = \(\frac{3}{X}\) = \(\sqrt{3}\)

x = \(\sqrt{3}\)

tan 45° = \(\frac{3}{y}\) = 1

y = 3

i.e, Width of river = n + y = 3 + \(\sqrt{3}\) m

11. A ladder 15 m long just reaches the top of a vertical wall. If the ladder makes an angle of 60° with the wall, the distance of the foot of the ladder from the wall is

    A. 7.5 m
    B. \(5 \sqrt{3}\) m
    C. \(10 \sqrt{3}\)m
    D. \(\frac{15 \sqrt{3}}{2}\) m


Answer – Option D

Explanation –


Let distance of foot of ladder from wall = x

sin 60° = \(\frac{x}{15}\) = \(\frac{\sqrt{3}}{2}\)

x = \(15 \frac{\sqrt{3}}{2}\)

12. From a point on the bridge across a river the angles of depression of the bank on the opposite
side of the river are 60° and 45° respectively. If the bridge is at a height of 4 m from the bank,
the width of the river is


    A. \( 4 \frac {3 + \sqrt{3}}{3}\) m
    B. \( 4 (3 + \sqrt{3})\) m
    C. \( 2 \frac{1 + \sqrt{3}}{3}\) m
    D. \( 2 (1 + 2 \sqrt{3})\) m


Answer – Option A

Explanation –


From the figure,

tan60º = \(\frac{4}{x}\) = \(\sqrt{3}\)

x = \(\frac{4}{\sqrt{3}}\)

tan45º = \(\frac{4}{y}\) = 1

y = 4

i.e, width of river = x + y

= \(\frac{4}{\sqrt{3}}\) + 4

= \( 4 \frac {3 + \sqrt{3}}{3}\) m

13. The angle of elevation of the top of a tower from a point on the ground Inch is 30 m away from the foot of the tower is 45°. The height of the tow t is

    A. 15 m
    B. 10 \(\sqrt{3}\) m
    C. 30 m
    D. 30 \(\sqrt{3}\) m


Answer – Option C

Explanation –


tan 45° = \(\frac{h}{30}\) = 1

h = 30 m

14. From a point on the bridge across a river the angle of depressions of the bank on the opposite
side of the river are 30° and 60° respectively. If the bridge is at a height of 3 m from the bank, the width of the river is


    A. 4 \(\sqrt{3}\) m
    B. 2 \((\sqrt{3} + 1)\) m
    C. 2 \((\sqrt{3} + 3)\) m
    D. 2 \(\sqrt{3} \) m


Answer – Option A

Explanation –


tan 60º = \(\frac{3}{x}\) = \(\sqrt{3}\)

x = \(\sqrt{3}\)

tan 45º = \(\frac{3}{y}\) = 1

y = 3

i.e, width of river = x + y = 4 \(\sqrt{3}\)

15. The angle of elevation of top of a tower from a point on the groin 20 m away from the foot of
the tower is 60°. The height of tower is


    A. \(20 \frac{\sqrt{3}}{3}\) m
    B. \(40 \frac{\sqrt{3}}{3}\) m
    C. 20 m
    D. 20\(\sqrt{3}\) m


Answer – Option D

Explanation –


tan 60º = \(\sqrt{3}\) = \(\frac{n}{20}\) =20 \(\sqrt{3}\)

1. The angle of elevation of a ladder leaning against a wall is 60° and the foot of the ladder is 7.5 m away from the wall. The length of the ladder is

    A. 15 m
    B. 14.86 m
    C. 15. 64 m
    D. 15.8 m


Answer – Option A

Explanation –

According the question:


cos 60° = \(\frac{BC}{AC}\)

\(\frac{1}{2}\) = \(\frac{7.5}{AC}\)

A = 15 m

2. If sinθ + cosθ =\(\sqrt{3}\), then the value of \(\frac{3}{4}\)(tanθ + cotθ ) is

    A. 1
    B. \(\frac{3}{4}\)
    C. \(\frac{3}{2}\)
    D. 3


Answer – Option B

Explanation –

\({sinθ + cosθ }^{2}\) = \({sin}^{2}θ + {cos}^{2}θ + 2cosθsinθ \) = 3

cosθ. sinθ = 1, since \({sin}^{2}θ + {cos}^{2}θ\) = 1

\(\frac{3}{4}\)(tanθ + cotθ) = \(\frac{3}{4}\)\(\frac{{sin}^{2}θ + {cos}^{2}θ}{cosθsinθ}\)

= \(\frac{3}{4}\)

3. A ship is approaching a light home, 100 m high above the sea-level. The angle of depression of the ship as observed from the top of the light home, changes from 30° to 45°. The distance, in m, traveled by the ship during the period of observation , in m , is

    A. 100(\(\sqrt{3} + 1\))
    B. 100(\(\sqrt{3} – 1\))
    C. (\(100 \sqrt{3} + 1\))
    D. (\(100 \sqrt{3} – 1\))


Answer – Option B

Explanation –


X = \(100 \sqrt{3} – 100\)

4. If \(\frac{secθ + tanθ }{secθ – tanθ}\) = \(\frac{5}{3}\) 0º < 0 < 90º then the value of cosecθ is

    A. 2
    B. \( \frac{\sqrt{15}}{4}\)
    C. \( \frac {4}{\sqrt{15}}\)
    D. 4


Answer – Option D

Explanation –

\(\frac{secθ + tanθ }{secθ – tanθ}\) = \(\frac{5}{3}\)

8 tanθ = 2 secθ

\(\frac{4 sinθ}{cosθ}\) = \(\frac{1}{cosθ}\) = sinθ = \(\frac{1}{4}\)

cosecθ = 4

5. The length of the shadow of a pole is 90 m, when the sun’s elevation is 30°.The length of the shadow of the pole is x meter when Sun’s elevation is 60°. The value of x is

    A. 15 \(\sqrt{3}\)
    B. 30
    C. 45
    D. 20 \(\sqrt{3}\)


Answer – Option B

Explanation –

tan 30º = \(\frac{h}{90}\) = h = \( \frac {90} {\sqrt{3}}\)

tan 60º = \(\frac{h}{x}\) = x = \( \frac {90} {\sqrt{3} * \sqrt{3}}\)

since tan 30º = \( \frac {1} {\sqrt{3}}\)

and tan 60º = \(\sqrt{3}\)

6. If 3 sinθ+ 4 cosθ = 5, then the value of 4 sinθ – 3cosθ is

    A. -5
    B. -2
    C. 1
    D. 0


Answer – Option D

Explanation –

3 sinθ+ 4 cosθ = 5

\(\frac{3}{5}\) sin θ+ \(\frac{4}{5}\) cos θ = 5,

comparing with sinθ . sin θ + cos θ – cos θ = 1

we get sin θ = \(\frac{3}{5}\)

cos θ = \(\frac{4}{5}\)

= 4 sinθ – 3 cosθ = 5

= 4 \(\frac{3}{5}\) – 3 \(\frac{4}{5}\) = 0

7. From the top of a tower 50 m high the angles of depression of the top and bottom of a pole are
found to be 45° and 60° respectively. The height of the pole, in metre, is


    A. 50 (\(\sqrt{3} – 1\))
    B. 50 (\(\sqrt{3} + 1\))
    C. 50(\(3 – \sqrt{3} \)) m
    D. 50(\(6 \sqrt{3} – 1\)) m


Answer – Option A

Explanation –


According the question:

tan 45º = 1 = \(\frac{x}{50 + h}\)

x = 50 + h

tan 60º = \(\frac{x}{50}\)

\(\sqrt{3}\) * 50 = 50 + h

h = 50 (\(\sqrt{3} – 1\))

since (tan 60º = \(\sqrt{3}\))

8. If x = (1 + cotθ – cosecθ)(1 + tanθ + secθ),0º < 0 < 90º then the value of x is

    A. -2
    B. -1
    C. 1
    D. 2


Answer – Option D

Explanation –

(1 + cotθ – cosecθ)(1 + tanθ + secθ)

\((1 + \frac{cosθ}{sinθ} – \frac{1}{sinθ})(1 + \frac{sinθ}{cosθ} – \frac{1}{cosθ})\)

= 2 + \(\frac{{sin}^{2}θ + {cos}^{2}θ – 1 }{sinθcosθ}\)

= 2 + 0 = 2

9. The angle of elevation of the top of an unfinished tower at a point 20m away from its base is 45°. How much higher must the tower be raised so that its angle of elevation of the top at the same point be 60°

    A. 20 \(\sqrt{3}\)
    B. 20\(\sqrt{3}\) + 1 m
    C. 20\((\sqrt{3} – 1)\) m
    D. \(\frac{20 \sqrt{3} }{3} + 1\) m


Answer – Option C

Explanation –


tan 45º = 1 = \(\frac{x}{20}\)

x = 20

tan 60º = \(\sqrt{3}\) = \(\frac{h + x}{20}\)

h = 20 \(\sqrt{3}\) – 20

= 20 \((\sqrt{3} – 1)\)

10. The expression \({tan}^{2}θ + {cot}^{2}θ – {sec}^{2}θ{cosec}^{2}θ \) is equal to

    A. 0
    B. 1
    C. -1
    D. -2


Answer – Option D

Explanation –

\({tan}^{2}θ + {cot}^{2}θ – {sec}^{2}θ{cosec}^{2}θ \)

put θ= 45°, since all ‘θ’ values should give same value of expression

= 1 + 1 – \((\sqrt{2}^{2})(\sqrt{2}^{2})\)

= 2 – 4 = -2

11. Two poles of equal height stand on either side of a roadway which is 100 m wide. At a point in the roadway between the poles, the elevation of the tops of the poles are 60°and 30°. the height of each pole, in meter is

    A. 25 \(\sqrt{3}\)
    B. 50 \(\sqrt {3}\)
    C. 25 \((\sqrt{3} – 1)\)
    D. 25 \((\sqrt{3} + 1)\)


Answer – Option A

Explanation –


tan 60º = \(\sqrt{3}\) = \(\frac{AB}{BC}\)

BC = \(\frac{X}{\sqrt{3}}\)

tan 30º = \(\frac {1} {\sqrt{3}}\) = \(\frac{ED}{CD}\)

CD = x \(\sqrt{3}\)

\(\frac{A}{Q}\) = \(\frac {x} {\sqrt{3}}\) + x\(\sqrt{3}\) = 100

x = 25 \(\sqrt{3}\)

12. If \({2sin}^{2}θ – 5sinθcosθ + {7cos}^{2}θ = 1 \)then possible values of tanθ are

    A. 2, 3
    B. 1, 3
    C. 2, \(\frac{5}{2}\)
    D. 3, 4


Answer – Option A

Explanation –

\({2sin}^{2}θ – 5sinθcosθ + {7cos}^{2}θ = 1 \)

Divide by \({cos}^{2}θ \)

= \({2 tan}^{2}θ – 5 tanθ + 4 \)

= \(\frac{{sin}^{2}θ + {cos}^{2}θ}{{cos}^{2}θ}\) = \({tan}^{2}θ + 1\)

= \({tan}^{2}θ – tanθ + 6\)

(tanθ – 3) (tanθ – 2) = 0

tanθ = 2 or 3

13. From the top of a tower 90 m high, the angles of depression of the top and bottom of a building
are observed to be 30° and 60° respectively. The height of the building, in m, is


    A. 40
    B. 45 \(\sqrt{3}\)
    C. 60
    D. 40 \(\sqrt{3}\)


Answer – Option C

Explanation –


tan 60º = \(\sqrt{3}\) = \(\frac{90}{BC}\)

= BC = \(\frac{90}{\sqrt{3}}\)

tan 30º = \(\frac{1}{\sqrt{3}}\) = \(\frac{90}{\frac {AE}{\sqrt{3}}}\)

AE = 30

i.e, BE = 60 m

14. If cosθ – sinθ = l and tanθ = m \({sec}^{2}θ\), then the value of \(\frac{{I}^{2} + 2m}{2}\) is

    A. \(\frac{1}{2}\)
    B. \(\frac{1}{4}\)
    C. \(\frac{2}{4}\)
    D. 1


Answer – Option A

Explanation –

cosθ– sinθ = l ; tanθ = m sec2θ

put θ = 45°

l = 0 ; m = \(\frac{1}{2}\)

\(\frac{1 + 2m}{2}\) = \(\frac{0 + 2 * \frac{1}{2}}{2}\) = \(\frac{1}{2}\)

15. A person is standing on the ground and flying a kite with a string of length 140 m at an angle of 30°. Another person is standing on the roof of a building 20 m high and is flying a kite at an angle of 45°. If both persons are on opposite sides of both the kites, the length (in m ) of the string that the second person must have so that the two kites meets,

    A. 70
    B. 60 \(\sqrt{2}\)
    C. 50 \(\sqrt{2}\)
    D. 50


Answer – Option C

Explanation –


sin 30° = \(\frac{1}{2}\) = \(\frac{AB}{AC}\)

AB = 70m

AF = (70 – 20) m = 50m

AE = 50 \(\sqrt{2}\)


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