# Railway Exam –Aptitude (Trigonometry) Problems

Railway Exam –Aptitude (Trigonometry) Problems Set-19:
Dear Readers, Here we have given from Aptitude Questions for Railway Exam 2016. Candidates those who are all preparing for these exams can use this material.

1).A person standing on the bank of river observes that the angle of elevation of the top of tree standing on the opposite bank is 60°. When he was 40 m away from the bank, he finds that the angle of elevation to be 30°. What is the height of the tree?
a)  36.64 m
b)  38.64 m
c)  42.64 m
d)  34.64 m

2).From the top of a cliff 90 m high, the angles of depression of the top and bottom of a tower are observed to 30° and 60° respectively. What is the height of the tower?
a)  45 m
b)  60 m
c)  75 m
d)  30 m

3).The least value of (4sec2 Ɵ + 9 cosec2 Ɵ) is
a)  1
b)  19
c)  25
d)  7

4).If 5cos Ɵ + 12sin Ɵ = 13, then tan Ɵ is equal to
a)  13/12
b)  12/13
c)  12/5
d)  5/12

5).A person standing on the bank of river observes that the angle of elevation of the top of tree standing on the opposite bank is 60°. When he was 40 m away from the bank, he finds that the angle of elevation to be 30°. What is the height of the tree?
a)  36.64 m
b)  38.64 m
c)  42.64 m
d)  34.64 m

6).If tan Ɵ + sin Ɵ = m and  tan Ɵ – sin Ɵ = n, then m2 – n2 is equal to
a)  4√mn
b)  √mn
c)  2√mn
d)  None of these
7).If tan Ɵ = 12/13, then (2sin Ɵ cos Ɵ) / (cos2 Ɵ – sin2 Ɵ) is equal to
a)  123/25
b)  312/25
c)  231/25
d)  192/25

8).The value of x, if x cot245° – sec260° + sin230° = 1/8, is
a)  31
b)  312
c)  31/8
d)  None of these

9).If √3 tan 2 Ɵ – 3 = 0. Then the angle Ɵ is
a)  30°
b)  60°
c)  90°
d)  None of these

10).If tan(A – B) = 1/√3 and tan(A + B) = √3, then the values of A and B are
a)  45°, 15°
b)  40°, 20°
c)  30°, 60°
d)  None of these

1). d) 2). b) 3). c) 4). c) 5). d) 6). a) 7). b) 8). c) 9). a) 10). a)

Solution:

1).   Let height of tree (PQ) be h m.
Then, AP = h cot 60°       …(i)
And BP = h cot 30°          …(ii)

AB = BP – AP = h(cot 30° – cot 60°)
h = 40 / [√3 – (1 / √3)]
= (40  ×√3) / (3 – 1) = 20√3
= 20 × 1.732 = 34.64

2).Let the height of tower (AB) be h m and distance between the edges of cliff and tower be x m.

According to figure, AP / PQ = cot 60°
x / 90 = 1 / √3
x = 30√3m
and BM / MQ = cot 30°
x / (90 – h) = √3
90 – h = x /√3 = (30√3) / √3
90 – h = 30, h = 90 – 30 = 60 m

3).(4sec2 Ɵ + 9 cosec2 Ɵ)
= 4 (1 + tan2 Ɵ) + 9 (1 + cot2 Ɵ)
= 13 + (4 tan2 Ɵ + 9 cot2 Ɵ)
AM ≥ GM
(4 tan2 Ɵ + 9 cot2 Ɵ) / 2    ≥    √(4 tan2 Ɵ . 9 cot2Ɵ)
(4 tan2 Ɵ + 9 cot2 Ɵ) = 2 × √36  ≥ 12
Minimum value of (4sec2 Ɵ + 9 cosec2 Ɵ) = 13 + 12 = 25

4).

5 cos Ɵ + 12 sin Ɵ = 13

(5 / 13) cos Ɵ + (12 / 13) sin Ɵ = 1
sin ɸ cos Ɵ + cos ɸ sin Ɵ = sin 90°
sin (⍬ +  ɸ ) = sin 90°
Ɵ +  ɸ = 90°
Ɵ = 90° –  ɸ
tan Ɵ = tan(90° –  ɸ) = cot ɸ = 12 / 5

5).Let height of tree(PQ) be h m.
Then, AP = h cot 60°       …(i)
And BP = h cot 30°          …(ii)

AB = BP – AP = h(cot 30° – cot 60°)
h = 40 / [√3 – (1 / √3)
= (40  ×√3) / (3 – 1) = 20√3
= 20 × 1.732 = 34.64

6).m2 – n2 = (tan⍬ + sin⍬)2 – (tan⍬ – sin⍬)2
= tan2⍬ + sin2⍬ + 2(tan⍬. sin⍬)  – [tan2⍬ + sin2⍬ – 2(tan⍬. sin⍬)]
= 4 tan⍬. sin⍬
m2– n2 = 4 (sin2⍬ / cos⍬)
Now, mn = (tan⍬ + sin⍬) . (tan⍬ – sin⍬)
= tan2⍬ – sin2
=( sin2⍬ / cos2⍬) – sin2
= (sin2⍬ – sin2⍬ . cos2⍬) / cos2
= (sin2⍬ (1 –  cos2⍬) / cos2
√mn = (sin2⍬ / cos⍬)
m2– n2 = 4√mn

7).We have, tan⍬ = 12 / 13
P = 12 and B = 13
So, H2= 122 + 132
= 144 + 169 = 313
H = √(313)
sin⍬  = 12 / √(313)
cos⍬ = 13 / √(313)
Now,
2sin⍬cos⍬ / (cos2⍬ – sin2⍬) = [ 2 × (12 / √313) × (13 / √313) ] / [(132 / 313) – (122 / 313)]
= 312 / (169 – 144) = 312 / 25

8).x cot2 45° – sec2 60° + sin2 30° = 1 / 8
x – (2)2 + (1 / 2)2 = 1 / 8
x – 4 + (1 / 4) = 1 / 8
x = (1 / 8) – (1 / 4) + 4 = (1 – 2 + 32) / 8 = 31 / 8

9).√3 tan 2 ⍬ – 3 = 0.
√3 tan 2 ⍬ = 3
tan 2 ⍬ = √3 = tan60°
2 ⍬ = 60°
⍬ = 30°

10).tan (A – B) = 1 / √3
tan (A – B) = tan30°
A – B = 30°                     ..(i)
And tan (A + B) = √3 =  tan60°
A + B = 60°                     ..(ii)
On solving Eqs. (i) and (ii) we get
A = 45         ° and B = 15° 