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1.

In ∆ABC measure of angle B is 90o. If secA = 25/7, and AB = 14 cm, then what is the length (in cm) of side BC?1). 502). 483). 204). 26

Answer»

<P>Perpendicular = AB

Base = BC

Hypotenuse = AC

secA= 25/7

Then cosA= 7/25

SinA= √{1-(7²/25²)}

SinA= 24/25

SinA = P/B

And AB is 14 ( given)

If P= 24 by the RATIO 

Then 24 is EQUAL to 14

And ratio of 7 is equal to (14/24)*7

Which is equal to 4.08
Discussion
2.

The angles of elevation of the top of a tower from two points A and B which are at a distance of 20 m and 12 m from the base of the tower and are in the same straight line. The angles are complementary to each other. The height of the tower is1). 15 m2). 15.29 m3). 15.49 m4). 20.45 m

Answer»
Discussion
3.

The angle of elevation from a point on the ground to the top of building and to top of chimney which is on building are x° and 45° and the height of building is ‘h’ meter. What is the height of chimney (in meter)?1). h cot x + h2). h cot x - h3). h tan x - h4). h tan x + h

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Discussion
4.

1). sin2x > \(\frac{1}{2}\) and cos2x> \(\frac{1}{2}\)2). sin2x > \(\frac{1}{2}\) and cos2x < \(\frac{1}{2}\)3). sin2x < \(\frac{1}{2}\) and cos2x < \(\frac{1}{2}\)4). sin2x, cos2x at least one is less than 1

Answer»

If 0 ≤ X ≤ $(\frac{{\rm{\PI }}}{2})$

Then 0 ≤ sinx ≤ 1

0 ≤ cosx ≤ 1

∴ (IV) is ALWAYS right

sinx increases from 0° to 90° in first QUADRANT and cosx decreases.
Discussion
5.

1). π/42). π/23). 3π/44). π

Answer»

TAN(A - B) = (tanA – TANB)/(1 + tanA tanB) = (1/2 + 1/3)/(1 – 1/2 × 1/3) = 1

A – B = π/4
Discussion
6.

1). 1 – cos2A2). 2sin2A3). sec2A4). cosec2A

Answer»

Put cot(A) = COS(A)/sin (A), TAN(A) = sin(A)/cos (A) and sec(A) = 1/cos(A)

⇒ [(1/cos(A)/{ cos(A)/sin (A) + sin(A)/cos (A) }]2 and using sin2(A) + cos2(A) = 1

⇒ sin2(A)

∴ its SIMPLIFIED FORM is 1 – cos2(A)
Discussion
7.

1). ½2). 3/43). 34). 3/2

Answer»

⇒ given: tan θ = (4/3)

⇒ LET (3sin θ + 2COS θ)/(3sin θ - 2cos θ) = x

⇒ Divide numerator and DENOMINATOR with COS θ, we get

⇒ (5 tan θ + 3)/(5 tan θ - 3) = x

⇒ (3 × (4/3) + 2)/[3 × (4/3) - 2] = x

⇒ ((12/3) + 2)/((12/3) - 2) = x

⇒ 18/6 = x

∴ x = 3
Discussion
8.

If tan 330° = x, then the value of x is?1). -1/√32). -√33). -1/24). -1/√2

Answer»

⇒ TAN (330°) = tan (360° – 30°)

⇒ tan (360° – X) = -tan x[? tan (360° - θ) = -tan θ]

⇒ tan (360° – 30°) = -tan 30°

⇒ -tan 30° = -1/√3
Discussion
9.

What is the value of (cosec 45° + √3/2) ?1). 5/62). (4 + √3)/2√33). (2√2 + √3)/24). 3√3/2

Answer»

∴ the VALUE of (COSEC 45° + √3/2) = √2 + √3/2 = (2√2 + √3)/2
Discussion
10.

If tan \(A\; = \;\frac{{1 - cosB}}{{sinB}},\) then what is \(\frac{{2tanA}}{{1 + {{\tan }^2}A}}\) equal to?1). \(\frac{{sinB}}{2}\)2). 2tanB3). sinB4). 4sinB

Answer»

$(tanA\; = \;\frac{{1 - cosB}}{{sinB}}\; = \;\frac{{2{{\SIN }^2}\frac{B}{2}}}{{2\sin \frac{B}{2}\COS \frac{B}{2}}}\; = \;\tan \frac{B}{2})$

(? cos2x = 1 – 2sin2x & sin2x = 2sinxcosx)

⇒ A = B/2

We know that,

sin 2x = 2tanx/(1 + tan2x)

$(\begin{array}{l}\frac{{2tanA}}{{1 + {{\tan }^2}A}}\; = \;\sin \left( {2\; \TIMES \;\frac{B}{2}} \right)\; = \;\sin B\\\therefore {\rm{\;}}\frac{{2tanA}}{{1 + {{\tan }^2}A}}\; = \;\sin B\end{array})$
Discussion
11.

What is the value of cosecA/√(cosec2A – 1)?1). sinA2). secA3). tanA4). cosA

Answer»

We, know 1 + COT2A = cosec2A

⇒ cosec2A – 1 = cot2A

$(\therefore {\rm{\;}}\frac{{COSECA}}{{\sqrt {cose{c^2}A - 1} }}\; = \;\frac{{cosecA}}{{\sqrt {{{\cot }^2}A} }}\; = \;\frac{{cosecA}}{{cotA}}\; = \;\frac{{SINA}}{{sinAcosA}}\; = \;\frac{1}{{cosA}}\; = \;secA\;)$

(? cosecA = 1/sinA )(? cotA = cosA/sinA) 
Discussion
12.

ΔDEF is right angled at E. If m∠F = 30°, then find the value of (cos D – 1/√2). 1). -1/22). (√6 – 1)/√33). (√2 – 2)/2√24). √3 – 2

Answer»

We KNOW, SUM of all the ANGLES of a triangle is 180.

⇒ ∠D + ∠E + ∠F = 180°

 ⇒ ∠D + 90 + 30 = 180°

⇒ ∠D = 60°

∴ (cosD – 1/√2) = cos 60° – 1/√2 = 1/2 – 1/√2 = (√2 – 2­­)/2√2
Discussion
13.

The value of \(\frac{1}{{\left( {1 + {{\tan }^2}\theta } \right)}} + \frac{1}{{\left( {1 + {{\cot }^2}\theta } \right)}}\) is1). 1/42). 13). 24). 1/2

Answer»

Given expression is,

$(\frac{1}{{\left( {1 + {{\TAN }^2}\theta } \right)}} + \frac{1}{{\left( {1 + {{\cot }^2}\theta } \right)}})$

Using the property: 1 + tan2θ = sec2θ and 1 + cot2θ = COSEC2θ

$(= \frac{1}{{se{c^2}\theta }} + \frac{1}{{COSE{c^2}\theta }})$

Using the property: sec θ = 1/cosθ and cosec θ = 1/sin θ

$(= \frac{1}{{\frac{1}{{co{s^2}\theta }}}} + \frac{1}{{\frac{1}{{SI{n^2}\theta }}}})$

= cos2θ + sin2θ

? sin2θ + cos2θ = 1

= 1
Discussion
14.

1). 13.72). 14.73). 13.54). 15.7

Answer»
Discussion
15.

A ladder of length 6 m makes an angle of 45° with the floor while leaning against one wall of a room. If the foot of the ladder is kept fi x ed on the floor and it is made to lean against the opposite wall of the room, it makes an angle of 60° with the floor. Find the distance between these two walls of the room.1). 10 m2). 9.25 m3). 6.85 m4). 7.23 m

Answer»
Discussion
16.

If \(\sin \theta = \sqrt {\frac{{{x^2} - {y^2}}}{{{x^2}+ {y^2}}}}\), then the value of \(\sqrt {({{\sec }^2}\theta {\rm} + {\rm}{{\tan }^2}\theta})\) is (where 0° < θ < 90°) :1). \(\frac{{{x^2}}}{{x\sqrt {{y^2} - {x^2}} }}\)2). 13). x/y4). y/x

Answer»

$(\sin \theta = \sqrt {\frac{{{x^2} - {y^2}}}{{{x^2}+ {y^2}}}} )$

We know that, sin θ = Perpendicular/HYPOTENUSE

where,

Perpendicular $(= {\rm}\sqrt {{x^2} - {y^2}} )$ and

Hypotenuse $(= {\rm}\sqrt {{x^2} + {y^2}} )$

We know according to Pythagoras THEOREM,

(Hypotenuse)2 = (Perpendicular)2 + (BASE)2

⇒ x2 + y2 = x2 – y2 + (Base)2

⇒ Base $(= {\rm}\sqrt {{x^2}+ {y^2} - {x^2} + {y^2}} = \sqrt 2 y)$ 

SEC θ = Hypotenuse/Base $(= {\rm}\left( {\sqrt {{x^2}+{y^2}}} \right)/\sqrt 2 y)$

tan θ = Perpendicular/Base $(= {\rm}\left( {\sqrt {{x^2} - {y^2}}} \right)/\sqrt 2 y)$

$(\sqrt {({{\sec }^2}\theta {\rm} + {\rm}{{\tan }^2}\theta })= \sqrt {\frac{{{x^2}+ {y^2}}}{{2{y^2}}}+ \frac{{{x^2} - {y^2}}}{{2{y^2}}}} = \frac{x}{y})$
Discussion
17.

1). 0.82). 1.253). 4/34). 3/4

Answer»

USING trigonometric identities,

sec2 θ = 1 + tan2 θ

⇒ tan2 θ = sec2 θ - 1 = 25/9 - 1 = 16/9

Also,

COSEC2 θ = 1 + cot2 θ = 1 + 1/tan2 θ = 1 + 9/16 = 25/16

∴ cosec θ = √(25/16) = 5/4 = 1.25
Discussion
18.

What is the least value of tan2 θ + cot2 θ + sin2 θ + cos2 θ + sec2 θ + cosec2 θ?1). 12). 53). 34). 7

Answer»

Using these trigonometric identities:

sin2 θ + cos2 θ = 1

tan2 θ + 1 = sec2 θ

cot2 θ + 1 = cosec2 θ

Required value: tan2 θ + cot2 θ + sin2 θ + cos2 θ + sec2 θ + cosec2 θ

On SIMPLIFYING we get:

⇒ tan2 θ + cot2 θ + 1 + 1 + tan2 θ + 1 + cot2 θ

⇒ 3 + 2(tan2 θ + cot2 θ)

Least value of tan2 θ + cot2 θ = 2

⇒ 3 + 2(2) =7
Discussion
19.

The angle of elevation of the top of a tower from two horizontal points (in opposite sides) at distances of x meter and x + 12 meter from the base of tower are 60° and 30° respectively. The value of x is1). 32). 63). 94). 12

Answer»

According to the situation,

Let the Height of the tower be h

tan 60° = h/x----(1)

h = x√3----(2)

tan 30° = h/(x + 12)----(3)

h = (x + 12)/√3----(4)

From (2) and (4)

x√3 = (x + 12) /√3

3x = x + 12

2x = 12

x = 6
Discussion
20.

What is the value of sin 75° + sin 15°?1). √3 2). 2√33). √(3/2)4). 3/√2

Answer»

⇒ SIN 15° + sin 75° = sin 15° + COS 15° = t

Squaring on both SIDES, we get

⇒ sin215° + cos215° + 2SIN 15°.cos 15° = t2

⇒ 1 + sin 30° = t2

⇒ 1 + 1/2 = t2

⇒ t = √(3/2)
Discussion
21.

1). 1/22). 1/√23). 1/√34). 2

Answer»
Discussion
22.

In a circle of radius 6 cm, an arc of certain length subtends 20° 17’ at the center. Find in sexagesimal unit the angle subtended by the same arc at the center of a circle of radius 8 cm1). 15° 12’ 45”2). 15° 11’ 45”3). 10° 12’ 45”4). 15° 12’ 40”

Answer»

Let an ARC of length be m cm subtends 20° 17’ at the center of a circle of radius 6 cm and α° at the center of a circle of radius 8 cm. 

Now, 20° 17’ = {20 (17/60)}° 

= (1217/60)°

= 1217π/(60 × 180) radian [since, 180° = π radian]

And α° = πα/180 radian

We KNOW, the formula, s = rθ then we GET,

When the circle of radius is 6 cm; m = 6 × [(1217π)/(60 × 180)] ………… (i) 

And when the circle of radius 8 cm; m = 8 × (πα)/180 …………… (II)

Therefore, from (i) and (ii) we get; 

8 × (πα)/180 = 6 × [(1217π)/(60 × 180)]

Or, α = [(6/8) × (1217/60)]° 

Or, α = (3/4) × 20° 17’ [since, (1217/60)° = 20° 17’]

Or, α = 3 × 5°4’ 15” 

Or, α = 15° 12’ 45”.
Discussion
23.

1). 20√32). 30(√3 + 1)3). 40(√3 - 1)4). 50(√3 - 1)

Answer»
Discussion
24.

ΔABC is right angled at B. If ∠A = 60°, then what is the value of Cot C?1). √22). 1/√33). √34). 2/√3

Answer»

SINCE ∠A = 60°, So ∠C = (90 - 60) = 30°

⇒ COT C = Cot 30° = √3

∴ Cot C = √3
Discussion
25.

If tan (2 θ + 45°) = cot 3 θ are acute angles, then the value of θ is1). 5°2). 9°3). 12°4). 15°

Answer»

We know that COT θ = tan (90° - θ)

In the GIVEN question,

tan (2 θ + 45°) = cot 3θ

⇒ tan (2 θ + 45°) = tan (90° - 3θ)

⇒ (2 θ + 45°) = N
Discussion
26.

What is the value of (cos 40° - cos 140°)/(sin 80° + sin 20°)?1). 2√32). 2/√33). 1/√34). √3

Answer»
Discussion
27.

From a point on a bridge across a river, the angles of depression of the banks are 30° and 45°, respectively. If the brige is at a height of 4 m above the bank, find the width of the river.1). 4(√3 + 1) m2). 4(√3 – 1) m3). (√3 – 1) m4). (√3 + 1) m

Answer»
Discussion
28.

1). 2 – √32).3).4).

Answer»

½ × AC × BD = ½ × AB × BC [Equating area]

AC × BD = AB × BC

4 BD × BD = AB × BC [∵ AC = 4 BD]

Dividing by BC2 both the sides

4 × BD/BC × BD/BD = AB/BC

In triangle BDC, sin C = BD/BC and In triangle ABC, tan C = AB/BC

4sin2C = tan C

4sin2C = sinC/cos C

4sinC cos C = 1

2sin C cos C = ½

Sin2C = ½ [∵ sin 2A = 2SINA cos A]

∴ 2C = 30° and C = 15°

tan C = tan 15°

To find tan 15°

Tan 2θ = 2tanθ/ (1 – tan2 θ)

LET tan 15° be x

Tan 30° = 2x/(1 – x2)

1/√3 = 2x/(1 – x2)

1 – x2 = 2√3x

x2 + 2√3x – 1 = 0

Using quadratic formula and ignoring the negative term

x = (-2√3 + 4)/2 = 2 – √3
Discussion
29.

1). \(2cotA + \;ta{n^2}A\)2). \(co{t^2}A - \;ta{n^2}A\)3). \(co{t^2}A - \;2tanA\)4). \(co{t^2}A + \;ta{n^2}A\)

Answer»

CONSIDER,

cosec A sec A COT A - sec A cosec A tan A

$(\RIGHTARROW \frac{1}{{\sin A}}\;\frac{1}{{cosA}}\;\frac{{cosA}}{{sinA}} - \frac{1}{{cosA}}\frac{1}{{sinA}}\frac{{sinA}}{{cosA}}\;\left[ {\because cosecA = \;\frac{1}{{sinA}};SECA = \frac{1}{{cosA}};tanA = \frac{{sinA}}{{cosA}};cotA = \frac{{cosA}}{{sinA}}} \right])$ 

$(\begin{array}{l} \Rightarrow \frac{1}{{si{n^2}A}} - \;\frac{1}{{co{s^2}A}}\\ \Rightarrow cose{c^2}A - \;se{c^2}A\end{array})$

$( \Rightarrow 1 + \;co{t^2}A - 1 - ta{n^2}A\;\left[ {\because1 + ta{n^2}A = \;se{c^2}A;1 + co{t^2}A = \;cose{c^2}A} \right])$ 

$(\Rightarrow co{t^2}A - \;ta{n^2}A)$
Discussion
30.

What is the simplified value of cosec 2A + cot 2A?1). sec A2). sec (A/2) 3). cot A4). cot 2A

Answer»

GIVEN: COSEC 2A + cot 2A

⇒ 1/sin 2A + cos 2A/sin 2A

⇒ (1 + cos 2A)/sin 2A…….(i)

We know that: TAN A = Sin 2A/(1 + cos 2A)…….(ii)

Here, (i) = 1/(ii)

So, REQUIRED value = 1/tan A = cot A
Discussion
31.

1). 88.242). 106.713). 92.154). 112.64

Answer»

RIGHT ANS is OPTION :3-92.15m
Discussion
32.

A tower 86 meter high is situated between two points A and B. the angle of elevation from point A is 60o and that of from B is 30o. the distance between two points is1). 345√3 m2). 344/√3 m3). 112√3 m4). 172/√3 m

Answer»
Discussion
33.

The distance between the two poles of length 16 m and 9 m, is X m. If two angles of elevation of their respective top from the bottom of the other are complementary to each other, then the value X is∶1). 10 m2). 15 m3). 16 m4). 12 m

Answer»
Discussion
34.

1). 7/252). 25/243). 8/254). 9/25

Answer»

GIVEN COT θ = 24/7

⇒ tan θ = 7/24

Since we know that SEC θ = √(1 + tan2 θ)

⇒ Sec θ = √(1 + 72/242)

⇒ Sec θ = √{(576 + 49)/242}

⇒ Sec θ = √{(252)/242} = 25/24

∴ Sec θ = 25/24
Discussion
35.

If sin 38° = x/y, then sec 38° - sin 52° is equal to1). \(\frac{{{x^2}}}{{y\sqrt {{y^2} - {x^2}} }}\)2). \(\frac{{{y^2}}}{{x\sqrt {{y^2} - {x^2}} }}\)3). \(\frac{{{x^2}}}{{y\sqrt {{x^2} - {y^2}} }}\)4). \(\frac{{{y^2}}}{{x\sqrt {{x^2} - {y^2}} }}\)

Answer»

cos 38° = √(1 – SIN238°) = √[1 – (X2/y2)] = √(y2 – x2)/y

sec 38° - sin 52°

= (1/cos 38°) - sin(90° - 38°)(? cos x = 1/sec x)

= (1/cos 38°) – cos 38°

= (1 – cos238°)/cos38°(? sin2x + cos2x = 1)

= sin238°/cos 38° 

$(\BEGIN{array}{L} = \;\frac{{\frac{{{x^2}}}{{{y^2}}}}}{{\frac{{\sqrt {{y^2} - {x^2}} }}{y}}}\; = \;\frac{{{x^2}}}{{y\sqrt {{y^2} - {x^2}} }}\\ \therefore {\rm{\;sec\;}}38^\circ {\rm{\;}} - {\rm{\;sin\;}}52^\circ \; = \;\frac{{{x^2}}}{{y\sqrt {{y^2} - {x^2}} }} \end{array})$
Discussion
36.

From the top of a cliff 90 metre high, the angles of depression of the top and bottom of a tower are observed to be 30° and 60° respectively. The height of the tower is :1). 45 m2). 60 m3). 75 m4). 30 m

Answer»
Discussion
37.

1). 02). 13). 24). 4

Answer»

sinA + sin B + sin C + sin D = 4

It is POSSIBLE when

sinA = sin B = sin C = sin D = 1

A = B = C = D = 90°

[sin 90° = 1]

Cos A + cos B + cos C + cos D = cos90° + cos90° + cos90° + cos90°

[cos90° = 0]

∴ 0
Discussion
38.

If [secA + tanA] / [secA – tanA] = 3, then find the value of [cosec2A + cot2A].1). 32). 93). 14). 7

Answer»

[secA + tanA] / [secA – tanA] = 3

⇒ secA + tanA = 3secA – 3tanA

⇒ 2secA = 4tanA

⇒ secA/tanA = 2

⇒ SINA = 1/2

∴ A = 30°

∴ cosec2A + cot2A = 4 + 3 = 7
Discussion
39.

A stone was placed 20√3 m in front of a building. A man, standing in a window of the building, observes that the angle of depression of the stone from the window is 30°. When he observes from another window of the building, the angle of depression becomes 60°. Find the distance between the two windows.1). 20 m2). 40 m3). 60 m4). 80 m

Answer»
Discussion
40.

1). 40(3 - √3) meter2). 30(4 - √3) meter3). 40√3(1 - √3) meter4). 40/(√3 - 1) meter

Answer»
Discussion
41.

If cosec2 θ = 625/576, then what is the value of [(sin θ – cos θ)/(sin θ + cos θ)]?1). 12). 31/173). 17/314). 14/25

Answer»

GIVEN, cosec2 θ = 625/576

⇒ SIN2 θ = 1/cosec2 θ = 576/625

⇒ sin θ = √(576/625) = 24/25

As we know, COS2 θ = 1 – sin2 θ = 1 – 576/625 = 49/625

⇒ cos θ = √(49/625) = 7/25

∴ [(sin θ – cos θ)/(sin θ + cos θ)] = (24/25 – 7/25)/(24/25 + 7/25) = (17/25)/(31/25) = 17/31
Discussion
42.

Heena observed from the top of the tower which is 80 m above the surface of river. The angle of elevation of the balloon in the sky measures 30 degree and the angle of depression of shadow of the balloon in the river is 60 degree. The height of the balloon above the surface of the river is?1). 402). 803). 1604). 320

Answer»
Discussion
43.

In ΔPQR measure of angle Q is 90°. If cotP = 8/15, and PQ = 4 cm, then what is the length (in cm) of side PR?1). 8.52). 7.53). 54). 4

Answer»

In ΔPQR measure of ANGLE Q is 90°,

COT P = Base/Perpendicalar

⇒ Cot P = PQ/QR

⇒ PQ/QR = 8/15

⇒ 4/QR = 8/15

⇒ QR = 7.5

By pythagoras Theorem,

PR2 = PQ2 + QR2

⇒ PR2 = 42 + 7.52

⇒ PR2 = 16 + 56.25 = 72.25

∴ PR = 8.5 cm
Discussion
44.

1). \(a\left[ {\frac{{{a^2} - {b^2}}}{{{a^2} + {b^2}}}} \right]\)2). \(b\left[ {\frac{{{a^2} + {b^2}}}{{{a^2} - {b^2}}}} \right]\)3). \(a\left[ {\frac{{{a^2} - {b^2}}}{{{a^2} + {b^2}}}} \right]\)4). \(a\left[ {\frac{{{a^2} + {b^2}}}{{{a^2} - {b^2}}}} \right]\)

Answer»
Discussion
45.

What is the simplified value of \(\frac{{cotA + tanB}}{{cotB + tanA}}\) ?1). tan B cot A2). tan A cot B3). tan A tan B4). cot A cot B

Answer»

$(\begin{ARRAY}{l}\frac{{cotA + TANB}}{{COTB + tanA}}\\ \Rightarrow \frac{{\frac{{cosA}}{{sinA}} + \frac{{SINB}}{{cosB}}}}{{\frac{{cosB}}{{sinB}} + \frac{{sinA}}{{cosA}}}}\\ \Rightarrow \frac{{cosA\;cosB + sinA\;sinB}}{{cosA\;cosB + sinA\;sinB}} \TIMES \frac{{sinB\;cosA}}{{sinA\;cosB}}\end{array})$

⇒ tanB cotA
Discussion
46.

What is the value of 4cos330° + 2sec2 60° + tan260°sin2 45° - sin30°tan245° - 4/cosec360°?1). 72). 83). 94). 10

Answer»

4cos330° + 2sec2 60° + tan260°sin2 45° - sin30°tan245° - 4/cosec360°

$(\; = \;4\; \TIMES \;\FRAC{{3\sqrt 3 }}{8} + 2\; \times \;4 + 3\; \times \;\frac{1}{2} - \frac{1}{2}\; \times \;1 - \frac{4}{{\frac{8}{{3\sqrt 3 }}}}\; = \;9)$

∴ 4cos330° + 2sec2 60° + tan260°sin2 45° - sin30°tan245° - 4/cosec360° = 9
Discussion
47.

If Sin θ – Cos θ = 0, then what is the value of (Sin4 θ + Cos4 θ + tan2 θ)?1). 22). 1.53). 14). 2.5

Answer»

Given that,

SIN θ – Cos θ = 0

⇒ Sin θ = Cos θ

When θ = 45°

So, Sin4 θ + Cos4 θ + tan2 θ

⇒ Sin4 45° + Cos4 45° + tan2 45°

⇒ (1/√2)4 + (1/√2)4 + 12

⇒ 1/4 + 1/4 + 1

⇒ 1.5
Discussion
48.

From a point 30 m away from the foot of a tower, the angle of elevation of the top of the tower is 30°. The height of the tower is1). 15√3 m2). 20√3 m3). 10√3 m4). 30√3 m

Answer»
Discussion
49.

What is the simplified value of \(\left[ {\frac{{\cos A}}{{(1 - \tan A)\;}} + {{\frac{{\sin A}}{{(1 - \cot A)\;}}}}} \right]^2?\)1). sin A + cos A2). 1 + sin 2A3). 1 + cos 2A4). tan A + cot A

Answer»

Put tan A = sin A/cos A and COT A = cos A/sin A 

⇒ $({\left( {\frac{{cosA}}{{1 - \frac{{SINA}}{{cosA}}}}\; + \;\frac{{sinA}}{{1 - \frac{{cosA}}{{sinA}}}}} \right)^2})$

⇒ $({\left( {\frac{{{{\cos }^2}A}}{{\cos A - \sin A}}\; - \;\frac{{{{\sin }^2}A}}{{\cos A - \sin A}}} \right)^2})$

⇒ $({\left( {\frac{{\left( {cosA - sinA} \right)\;\left( {\cos A\; + \;\sin A} \right)\;}}{{\cos A - \sin A}}} \right)^2})$

⇒ (cosA + sinA) 2

⇒ cos2A + sin2A + 2sinA cos A, Since we know that cos2A + cos2A = 1 and 2sinA cosA = SIN2A

⇒ 1 + sin2A

∴ its simplified value is 1 + sin2A
Discussion
50.

A ladder leans against a vertical wall. Top of the ladder is 5 m above the ground. When the bottom of the ladder is moved 1 m further away from the wall, the top of the ladder rest against the foot of the wall. What is the length of the ladder?1). 13 m2). 15 m3). 17 m4). 20 m

Answer»
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