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 \(\sqrt{2}\)