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?]
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:
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:
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:
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:
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
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:
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:
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:
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:
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
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 – \(\frac{{sin}^{2}y}{1 + cos y} + \frac{1 + cos y}{sin y} – \frac {sin}{1 – cos y}is equal to:\)
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:
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:
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
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
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
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?
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
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:
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
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
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
Answer – Option B
Explanation –
Answer – Option B
Explanation –
Answer – Option D
Explanation –
Answer – Option D
Explanation –
Answer – Option A
Explanation –
Answer – Option C
Explanation –
Answer – Option A
Explanation –
Answer – Option D
Explanation –
Answer – Option A
Explanation –
According the question:
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
Answer – Option B
Explanation –
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
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
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
Answer – Option A
Explanation –
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°
Answer – Option C
Explanation –
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
Answer – Option A
Explanation –
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
Answer – Option C
Explanation –
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,
Answer – Option C
Explanation –