Trigonometry: SSC CGL Practice Questions

Ans: D

Explanation: This problem uses trigonometry. The ladder, the building, and the ground form a right triangle. The ladder is the hypotenuse (5 meters), the height the ladder reaches on the building is the opposite side to the 30-degree angle, and the distance on the ground is the adjacent side. We use the sine function: sin(angle) = opposite/hypotenuse. So, sin(30 degrees) = height / 5. sin(30) = 0.5. Therefore, height = 5 * 0.5 = 2.5 meters.

Ans: D

Explanation: Let h be the height of the building and x be the horizontal distance between the building and the tower.
From the top of the building, tan(30) = (72-h)/x. Since tan(30) = 1/sqrt(3), we have x = (72-h)sqrt(3).
From the bottom of the building, tan(60) = (72-h)/x. However, this is incorrect. The angle of elevation from the *bottom* of the building to the top of the tower includes the height of the building. Let y be the horizontal distance between the building and the tower. We have,
tan(60) = (72-h)/x is incorrect. Let the distance between tower and building be ‘y’
We have two right angled triangles.
Consider from the top of the building:
tan 30 = (72 – h) / y
1/sqrt(3) = (72 – h) / y
y = (72 – h) * sqrt(3) —- (1)
Consider from the bottom of the building. The angle is 60.
tan 60 = (72 – h) / y.
The correct approach. From the top, let the horizontal distance be y. tan 30 = (72 – h) / y. y = (72 – h) / tan 30 = (72-h)*sqrt(3). From the bottom: Height of tower is 72.
From the bottom the building to the top of the tower, this angle is 60, not an angle of depression.
So tan(60) = (72-0)/x, which is incorrect.
The diagram is as follows: The building has height ‘h’. The tower has height 72. Distance between them is ‘x’.
tan(30) = (72-h)/x
tan(60) = 72-h /x +h – 0. Let h = building height. Then, tan 60 = h+ (72-h)/x.
From the bottom to the top of tower is actually incorrect.
Let’s redraw.
Let the building have height h. Tower height is 72.
Let distance between building and tower be x.
From top of building, angle is 30.
tan(30) = (72-h) / x.
x = (72-h) / tan(30) = (72-h) * sqrt(3) — (1)
At the bottom, the angle should be 60. But instead this says the angle to the top of the tower from the bottom of the building is 60.
tan(60) = (72-h) / x — WRONG
But it should be tan(60) = some vertical distance to the top / horizontal distance.
It should be:
Consider from bottom, and angle is 60
tan(60) = (72 – h) / x is not correct.
Let h be the height of the building. Let x be the distance between building and tower.
From the top of the building. The vertical distance to be observed is (72-h).
tan(30) = (72-h)/x, then x = (72-h)/tan(30)
From the bottom of the building, and viewing tower to the top, it should be the total height.
tan(60) = 72/x. x = 72 / tan(60)
x = (72 – h) / tan(30)
x = 72 / tan(60)
Therefore, (72-h)/tan(30) = 72 / tan(60)
(72-h) / (1/sqrt(3)) = 72/sqrt(3)
(72-h) * sqrt(3) = 72/sqrt(3)
72-h = 72/sqrt(3) / sqrt(3) = 72/3 = 24
h = 72 – 24 = 48

Correct Option: D

Ans: C

Explanation: We can use trigonometry to solve this problem. The tree and its shadow form a right triangle. The height of the tree is the opposite side, and the length of the shadow is the adjacent side to the angle we want to find. We can use the tangent function: tan(angle) = opposite/adjacent. In this case, tan(angle) = 30 / (10√3) = 3 / √3 = √3. The angle whose tangent is √3 is 60 degrees.
Correct Option: C

Ans: A

Explanation: We can use trigonometric identities to simplify the expression.
Recall that:
1. csc(x) = 1/sin(x)
2. sec(x) = 1/cos(x)
3. csc(90 – x) = sec(x)
4. sec(90 – x) = csc(x)

The given expression is: csc(60° + A) – sec(30° – A) + csc(49°) / sec(41°)

Let’s simplify csc(60° + A) – sec(30° – A). We can rewrite sec(30° – A) as csc(90° – (30° – A)) = csc(60° + A). Therefore, csc(60° + A) – sec(30° – A) = csc(60° + A) – csc(60° + A) = 0.

Now, let’s simplify csc(49°) / sec(41°). We can rewrite sec(41°) as csc(90° – 41°) = csc(49°). Therefore, csc(49°) / sec(41°) = csc(49°) / csc(49°) = 1.

So, the original expression simplifies to: 0 + 1 = 1.

Correct Option: A

Ans: B

Explanation:
We can use the identity: tan(x) * tan(90-x) = 1.
We can rearrange the terms:
tan 10 * tan 80 = tan 10 * tan (90-10) = tan 10 * cot 10 = 1
tan 20 * tan 70 = tan 20 * tan (90-20) = tan 20 * cot 20 = 1
tan 30 * tan 60 = (1/sqrt(3)) * sqrt(3) = 1
tan 40 * tan 50 = tan 40 * tan (90-40) = tan 40 * cot 40 = 1

Therefore the expression is:
(tan 10 * tan 80) * (tan 20 * tan 70) * (tan 30 * tan 60) * (tan 40 * tan 50)
= 1 * 1 * 1 * 1 = 1

Ans: A

Explanation: We are given cos(θ) = 35/37. We need to find csc(θ). First, find sin(θ) using the Pythagorean identity: sin²(θ) + cos²(θ) = 1. Therefore, sin²(θ) = 1 – cos²(θ) = 1 – (35/37)² = 1 – 1225/1369 = (1369 – 1225)/1369 = 144/1369. Taking the square root gives sin(θ) = √(144/1369) = 12/37. Then, csc(θ) = 1/sin(θ) = 1 / (12/37) = 37/12.

Correct Option: A

Ans: C

Explanation:
Let sin(θ) = s and cos(θ) = c.
We are given that s⁶ + c⁶ = 1/3.
We know that s² + c² = 1.
We want to find the value of s*c.

We can rewrite s⁶ + c⁶ as (s²)³ + (c²)³.
Using the identity a³ + b³ = (a + b)(a² – ab + b²), we have:
s⁶ + c⁶ = (s² + c²)(s⁴ – s²c² + c⁴) = 1 * (s⁴ – s²c² + c⁴)
Since s⁶ + c⁶ = 1/3, we get:
s⁴ – s²c² + c⁴ = 1/3
Also, (s² + c²)² = s⁴ + 2s²c² + c⁴ = 1² = 1.
We can rewrite the first equation as:
(s⁴ + 2s²c² + c⁴) – 3s²c² = 1/3
1 – 3s²c² = 1/3
3s²c² = 1 – 1/3 = 2/3
s²c² = 2/9
sc = √(2/9) = √2 / 3

Correct Option: C

Ans: D

Explanation: We are given the equation $3 \sin^2 \theta – \cos \theta – 1 = 0$. We can rewrite this equation using the identity $\sin^2 \theta = 1 – \cos^2 \theta$. Substituting, we have:
$3(1 – \cos^2 \theta) – \cos \theta – 1 = 0$
$3 – 3\cos^2 \theta – \cos \theta – 1 = 0$
$3\cos^2 \theta + \cos \theta – 2 = 0$
Let $x = \cos \theta$. Then we have the quadratic equation $3x^2 + x – 2 = 0$.
We can factor this quadratic as $(3x – 2)(x + 1) = 0$.
Thus, the possible values for x (which is $\cos \theta$) are $x = \frac{2}{3}$ or $x = -1$.
Since $0 < \theta < 90^\circ$, we know that $0 < \cos \theta < 1$. Therefore, we must have $\cos \theta = \frac{2}{3}$.
Now, we can find $\sin \theta$ using the identity $\sin^2 \theta + \cos^2 \theta = 1$:
$\sin^2 \theta = 1 – \cos^2 \theta = 1 – \left(\frac{2}{3}\right)^2 = 1 – \frac{4}{9} = \frac{5}{9}$
Since $0 < \theta < 90^\circ$, we have $\sin \theta = \sqrt{\frac{5}{9}} = \frac{\sqrt{5}}{3}$.
Now, we can find $\cot \theta$ and $\csc \theta$:
$\cot \theta = \frac{\cos \theta}{\sin \theta} = \frac{\frac{2}{3}}{\frac{\sqrt{5}}{3}} = \frac{2}{\sqrt{5}}$
$\csc \theta = \frac{1}{\sin \theta} = \frac{1}{\frac{\sqrt{5}}{3}} = \frac{3}{\sqrt{5}}$
Finally, we can find $\cot \theta + \csc \theta$:
$\cot \theta + \csc \theta = \frac{2}{\sqrt{5}} + \frac{3}{\sqrt{5}} = \frac{5}{\sqrt{5}} = \sqrt{5}$

Correct Option: D

Ans: A

Explanation:
We are given $3 \tan\theta = 2\sqrt{3} \sin\theta$. Since $\tan\theta = \frac{\sin\theta}{\cos\theta}$, we can substitute and obtain $3 \frac{\sin\theta}{\cos\theta} = 2\sqrt{3} \sin\theta$.
Case 1: $\sin\theta = 0$. This means $\theta = 0$, which is within the given range.
Case 2: $\sin\theta \ne 0$. Then we can divide both sides by $ \sin\theta$ which simplifies to $3/\cos\theta = 2\sqrt{3}$. Therefore $\cos\theta = \frac{3}{2\sqrt{3}} = \frac{\sqrt{3}}{2}$. This gives $\theta = 30^\circ$.
Now, evaluate the given expression for both cases:
Case 1: $\theta = 0$. This makes $\sin\theta = 0$, $\tan\theta = 0$. Then $2\theta = 0$, so $\sin 2\theta = 0$, $\cos 2\theta = 1$, $\tan 2\theta = 0$. The expression becomes undefined because $\csc 2\theta = \frac{1}{\sin 2\theta}$. Hence, we disregard this case.
Case 2: $\theta = 30^\circ$. Then $2\theta = 60^\circ$. So:
$\sin \theta = \frac{1}{2}$, $\cos \theta = \frac{\sqrt{3}}{2}$.
$\sin 2\theta = \sin 60^\circ = \frac{\sqrt{3}}{2}$, $\cos 2\theta = \frac{1}{2}$, $\tan 2\theta = \sqrt{3}$.
$\csc 2\theta = \frac{2}{\sqrt{3}}$, $\cot 2\theta = \frac{1}{\sqrt{3}}$.
$\sin^2 \theta = \frac{1}{4}$
$\tan^2 2\theta = 3$
So, the given expression is $\frac{\csc^2 2\theta + \cot^2 2\theta}{\sin^2 \theta + \tan^2 2\theta} = \frac{\frac{4}{3} + \frac{1}{3}}{\frac{1}{4} + 3} = \frac{\frac{5}{3}}{\frac{13}{4}} = \frac{5}{3} \cdot \frac{4}{13} = \frac{20}{39}$.

Correct Option: A

Ans: A

Explanation:
We are given that $\sin \theta = \frac{2ab}{a^2 + b^2}$. We need to find $\tan \theta$.
We know that $\sin^2 \theta + \cos^2 \theta = 1$.
So, $\cos^2 \theta = 1 – \sin^2 \theta$.
Substituting the given value of $\sin \theta$, we have:
$\cos^2 \theta = 1 – \left(\frac{2ab}{a^2 + b^2}\right)^2$
$\cos^2 \theta = 1 – \frac{4a^2b^2}{(a^2 + b^2)^2}$
$\cos^2 \theta = \frac{(a^2 + b^2)^2 – 4a^2b^2}{(a^2 + b^2)^2}$
$\cos^2 \theta = \frac{a^4 + 2a^2b^2 + b^4 – 4a^2b^2}{(a^2 + b^2)^2}$
$\cos^2 \theta = \frac{a^4 – 2a^2b^2 + b^4}{(a^2 + b^2)^2}$
$\cos^2 \theta = \frac{(a^2 – b^2)^2}{(a^2 + b^2)^2}$
Since $0^\circ \le \theta \le 90^\circ$, we have $\cos \theta = \frac{a^2 – b^2}{a^2 + b^2}$.
Now, we find $\tan \theta = \frac{\sin \theta}{\cos \theta}$.
$\tan \theta = \frac{\frac{2ab}{a^2 + b^2}}{\frac{a^2 – b^2}{a^2 + b^2}}$
$\tan \theta = \frac{2ab}{a^2 – b^2}$

Ans: B

Explanation: Let’s simplify the first part of the expression: ((sin(x+y) – 2sin(x) + sin(x-y))/(cos(x-y) + cos(x+y) – 2cos(x))). We can use the sum-to-product formulas:
sin(x+y) + sin(x-y) = 2sin(x)cos(y)
cos(x-y) + cos(x+y) = 2cos(x)cos(y)

So, the numerator becomes: 2sin(x)cos(y) – 2sin(x) = 2sin(x)(cos(y) – 1)
And the denominator becomes: 2cos(x)cos(y) – 2cos(x) = 2cos(x)(cos(y) – 1)

Therefore, the first part simplifies to: (2sin(x)(cos(y)-1)) / (2cos(x)(cos(y)-1)) = tan(x), if cos(y) != 1.

Now, let’s simplify the second part: ((sin(10x) – sin(8x))/(cos(10x) + cos(8x))). Using sum-to-product formulas:
sin(10x) – sin(8x) = 2cos((10x+8x)/2)sin((10x-8x)/2) = 2cos(9x)sin(x)
cos(10x) + cos(8x) = 2cos((10x+8x)/2)cos((10x-8x)/2) = 2cos(9x)cos(x)

Therefore, the second part simplifies to: (2cos(9x)sin(x)) / (2cos(9x)cos(x)) = tan(x), if cos(9x) != 0.

So the entire expression simplifies to: tan(x) * tan(x) = tan^2(x).

Correct Option: B

Ans: B

Explanation: We can use trigonometric identities to simplify this expression.

First, use the identity cos(A) – cos(B) = -2sin((A+B)/2)sin((A-B)/2) on the numerator:
cos 40° – cos 140° = -2sin((40+140)/2)sin((40-140)/2) = -2sin(90°)sin(-50°) = -2(1)(-sin 50°) = 2sin 50°

Next, use the identity sin(A) + sin(B) = 2sin((A+B)/2)cos((A-B)/2) on the denominator:
sin 80° + sin 20° = 2sin((80+20)/2)cos((80-20)/2) = 2sin(50°)cos(30°)

Now the original expression becomes:
(2sin 50°) / (2sin 50°cos 30°)

Simplifying, we get:
1 / cos 30°

Since cos 30° = √3/2, the expression equals:
1 / (√3/2) = 2/√3

Correct Option: B

Ans: B

Explanation: Let’s simplify the given expression step-by-step using trigonometric identities.
(tan θ + cot θ)(sec θ + tan θ)(1 – sin θ)

1. Express tan θ and cot θ in terms of sin θ and cos θ:
tan θ = sin θ / cos θ
cot θ = cos θ / sin θ
sec θ = 1 / cos θ

2. Substitute the expressions in the given expression:
[(sin θ / cos θ) + (cos θ / sin θ)] * [(1 / cos θ) + (sin θ / cos θ)] * (1 – sin θ)

3. Simplify the terms inside the brackets:
[(sin² θ + cos² θ) / (cos θ * sin θ)] * [(1 + sin θ) / cos θ] * (1 – sin θ)
Since sin² θ + cos² θ = 1,
[1 / (cos θ * sin θ)] * [(1 + sin θ) / cos θ] * (1 – sin θ)

4. Combine terms:
[(1 + sin θ)(1 – sin θ)] / [cos² θ * sin θ]
Using (a+b)(a-b) = a² – b²
(1 – sin² θ) / (cos² θ * sin θ)

5. Replace (1 – sin² θ) with cos² θ:
cos² θ / (cos² θ * sin θ)

6. Simplify:
1 / sin θ

7. Replace 1 / sin θ with cosec θ:
cosec θ

The correct answer should be cosec θ.

Correct Option: B

Ans: C

Explanation: Let ‘h’ be the height of the building. Let ‘x’ be the initial distance from the point on the ground to the building.

From the first scenario (angle of elevation = 45°):
tan(45°) = h/x
1 = h/x
x = h

From the second scenario (angle of elevation = 30°):
The distance is increased by 100 meters, so the new distance is x + 100
tan(30°) = h/(x + 100)
1/√3 = h/(x + 100)
x + 100 = h√3

Since x = h, substitute h for x:
h + 100 = h√3
100 = h√3 – h
100 = h(√3 – 1)
h = 100 / (√3 – 1)
Multiply the numerator and denominator by the conjugate (√3 + 1):
h = 100(√3 + 1) / ((√3 – 1)(√3 + 1))
h = 100(√3 + 1) / (3 – 1)
h = 100(√3 + 1) / 2
h = 50(√3 + 1)
h = 50√3 + 50

Ans: C

Explanation:
We can use trigonometric identities to solve this problem.

First, recall the identity: cos(90 – x) = sin(x).
Therefore, cos(75°) = cos(90° – 15°) = sin(15°).

Now we can rewrite the expression:
cos²(15°) + cos²(75°) + cot²(45°) = cos²(15°) + sin²(15°) + cot²(45°)

Using the identity cos²(x) + sin²(x) = 1, we know cos²(15°) + sin²(15°) = 1.

Also, we know that cot(45°) = 1. So, cot²(45°) = 1² = 1.

Therefore, the original expression simplifies to:
1 + 1 = 2

Ans: D

Explanation: Since triangle DEF is a right-angled triangle with the right angle at E, we know that angle D + angle F = 90 degrees. Given that angle F is 45 degrees, angle D must also be 45 degrees. Thus, triangle DEF is a 45-45-90 triangle. We have: csc(F) = csc(45°) = 1/sin(45°) = 1/(1/√2) = √2. cot(D) = cot(45°) = 1. Therefore, csc(F) * cot(D) = √2 * 1 = √2.

Correct Option: D

Ans: B

Explanation: This question utilizes the fundamental trigonometric identity: cos²θ + sin²θ = 1. In this case, θ = 60°. Therefore, cos²60° + sin²60° = 1.