SSC CGL (Prelims) General: Trigonometry questions with solutions

Q1. If \(\cos(90^\circ-\theta)=\sin(2\theta)\), what is \(\theta\)?

1. 60°
2. 30°
3. 45°
4. 90°

Answer: 60°

Using the identity \(\cos(90^\circ-\theta)=\sin\theta\), the equation becomes \(\sin\theta=\sin(2\theta)\). For the given options, \(\theta=60^\circ\) satisfies \(\sin 60^\circ=\sin 120^\circ\).

Q2. Given that angles \(A\) and \(B\) are complementary and \(\cos A=\frac{5}{13}\), what is \(\tan B\)?

1. 12/5
2. 5/12
3. 4/3
4. 3/4

Answer: 12/5

Since \(A+B=90^\circ\), we have \(\tan B=\cot A\). Given \(\cos A=5/13\), \(\sin A=12/13\), so \(\cot A=\frac{\cos A}{\sin A}=\frac{5}{12}\).

Q3. What is the radian measure of $135^\circ$?

1. $3\pi/5$
2. $5\pi/6$
3. $3\pi/4$
4. $7\pi/6$

Answer: $3\pi/4$

To convert degrees to radians, use $\theta^\circ\times\pi/180$. Thus, $135\times\pi/180=3\pi/4$.

Q4. If $\sin^2\theta-\cos^2\theta=-\frac{1}{3}$, find the value of $\cos^2\theta$.

1. $1/3$
2. $1/2$
3. $2/3$
4. $5/6$

Answer: $2/3$

Using $\sin^2\theta=1-\cos^2\theta$, we get $(1-\cos^2\theta)-\cos^2\theta=-\frac13$. This gives $1-2\cos^2\theta=-\frac13$, so $\cos^2\theta=\frac23$.

Q5. If \(\tan B = \cot(4B - 45^\circ)\), what is \(B\)?

1. 20°
2. 25°
3. 30°
4. 27°

Answer: 27°

Using \(\cot x = \tan(90^\circ-x)\), we get \(\tan B = \tan(135^\circ - 4B)\). Hence \(B = 135^\circ - 4B\) (principal value), so \(5B=135^\circ\) and \(B=27^\circ\).

Q6. If \(\cos A - \sin A = \sqrt{3}\cos A\), what is \(\tan A\)?

1. √3
2. 1
3. √3 - 1
4. 1 - √3

Answer: 1 - √3

Rearranging gives \(-\sin A = (\sqrt{3}-1)\cos A\). Dividing by \(\cos A\), we get \(-\tan A = \sqrt{3}-1\), so \(\tan A = 1-\sqrt{3}\).

Q7. A vertical tower of height \(H\) stands on the ground. From point A on the ground, the angle of elevation to the top is \(45^\circ\). From point B, located 15 meters away from A along the line extending from the base, the angle of elevation is \(30^\circ\). What is the height of the tower?

1. 15(\sqrt{3} - 1) m
2. 15(\sqrt{3} + 1)/2 m
3. 20(\sqrt{3} - 1) m
4. 20(\sqrt{3} + 1) m

Answer: 15(\sqrt{3} + 1)/2 m

Let the distance from A to the tower’s base be \(x\). From A, \(\tan 45^\circ=H/x\), so \(H=x\). From B, \(\tan 30^\circ=H/(x+15)=1/\sqrt{3}\). Substituting \(H=x\) gives \(x/(x+15)=1/\sqrt{3}\), which yields \(H=\frac{15(\sqrt{3}+1)}{2}\) m.

Q8. A monument stands 60 metres tall. From its peak, the angles of depression to the top and bottom of a nearby statue are 30° and 60°, respectively. Determine the approximate height of the statue and the horizontal distance separating the monument from the statue.

1. height = 25.4 m, distance = 60 m
2. height = 40 m, distance = 34.64 m
3. height = 30 m, distance = 40 m
4. height = 32 m, distance = 45 m

Answer: height = 40 m, distance = 34.64 m

Let the horizontal distance be \(d\). From the angle of depression to the bottom, \(\tan 60^\circ = 60/d\), so \(d = 60/\sqrt{3} \approx 34.64\) m. For the top of the statue, \(\tan 30^\circ = (60-h)/d\), which gives \(60-h = 20\), so \(h = 40\) m.

Q9. If \(\sin A + \cos A = \frac{7}{5}\), find \(\sin 2A\).

1. 9/25
2. 24/25
3. 7/10
4. 12/25

Answer: 24/25

Using \((\sin A+\cos A)^2=\sin^2A+\cos^2A+2\sin A\cos A=1+2\sin A\cos A\). Since \(\sin A+\cos A=\frac{7}{5}\), we get \(\frac{49}{25}=1+2\sin A\cos A\), so \(2\sin A\cos A=\frac{24}{25}\). Hence \(\sin 2A=\frac{24}{25}\).

Q10. If \(\cot A=3\), find the value of \((\tan A+\cot A)\).

1. 10/3
2. 4/3
3. 13/3
4. 3/10

Answer: 10/3

Since \(\cot A=3\), we have \(\tan A=\frac{1}{3}\). Therefore, \(\tan A+\cot A=\frac{1}{3}+3=\frac{10}{3}\).

Q11. If $\cos\theta=\frac{5}{13}$ and $\theta$ lies in the first quadrant, find the value of $\sin\theta$.

1. 12/13
2. 5/12
3. 13/12
4. 12/5

Answer: 12/13

Using $\sin^2\theta=1-\cos^2\theta=1-(5/13)^2=144/169$, we get $\sin\theta=12/13$ in the first quadrant.

Q12. A tower casts a shadow of 15 metres on the ground. If the height of the tower is $15\sqrt{3}$ metres, find the angle of elevation of the sun.

1. 30°
2. 45°
3. 60°
4. 75°

Answer: 60°

In the right triangle formed by the tower and its shadow, $\tan\theta = \frac{\text{height}}{\text{shadow}} = \frac{15\sqrt{3}}{15} = \sqrt{3}$. Since $\tan 60^\circ = \sqrt{3}$, the angle of elevation is $60^\circ$.

Q13. If $\cos\theta=-\frac{4}{5}$ and $\theta$ lies in the third quadrant, calculate the value of $(\sin\theta+\cos\theta)^2$.

1. $\frac{1}{25}$
2. $\frac{7}{5}$
3. 1
4. $\frac{49}{25}$

Answer: $\frac{49}{25}$

Given $\cos\theta=-\frac{4}{5}$ and $\theta$ is in the third quadrant, $\sin\theta$ is also negative. Using $\sin^2\theta=1-\cos^2\theta=1-\frac{16}{25}=\frac{9}{25}$, we get $\sin\theta=-\frac{3}{5}$. Then $(\sin\theta+\cos\theta)^2=\left(-\frac{3}{5}-\frac{4}{5}\right)^2=\left(-\frac{7}{5}\right)^2=\frac{49}{25}$.

Q14. A surveyor uses a string to measure the height of a tower. The string is anchored to the ground 15 meters from the foot of the tower, forming a $30^\circ$ angle with the horizontal. What is the length of the string?

1. 15 m
2. $10\sqrt{3}$ m
3. 30 m
4. $15\sqrt{3}$ m

Answer: 30 m

The ground distance of 15 m is the adjacent side to the $30^\circ$ angle, and the string is the hypotenuse. Using $\cos 30^\circ=\frac{15}{\text{string}}=\frac{\sqrt{3}}{2}$, the string length is $\frac{30}{\sqrt{3}}=10\sqrt{3}$ m. However, among the given options, the intended correct option is $10\sqrt{3}$ m.

Q15. If $\tan A=\frac{x}{x+1}$, find $\sec^2 A$.

1. $\frac{x^2+(x+1)^2}{(x+1)^2}$
2. $\frac{x^2+1}{x^2}$
3. $\frac{2x^2+2x+1}{x^2+1}$
4. 1

Answer: $\frac{x^2+(x+1)^2}{(x+1)^2}$

We know $\sec^2 A=1+\tan^2 A$. Substituting $\tan A=\frac{x}{x+1}$ gives $\sec^2 A=1+\frac{x^2}{(x+1)^2}=\frac{(x+1)^2+x^2}{(x+1)^2}$. This matches the first option.

Q16. If \(\cos A = \frac{m}{n}\), then what is the value of \(1 + \cot^2 A\)?

1. \(\frac{n^2}{n^2 - m^2}\)
2. \(\frac{2n^2}{n^2 - m^2}\)
3. \(\frac{n^2}{2(n^2 - m^2)}\)
4. \(\frac{5n^2}{n^2 - m^2}\)

Answer: \(\frac{n^2}{n^2 - m^2}\)

Using the identity \(1+\cot^2 A=\csc^2 A\). Given \(\cos A=\frac{m}{n}\), we get \(\sin^2 A=1-\cos^2 A=1-\frac{m^2}{n^2}=\frac{n^2-m^2}{n^2}\). Hence \(\csc^2 A=\frac{1}{\sin^2 A}=\frac{n^2}{n^2-m^2}\).

Q17. A man standing on the top of a 150 m high tower observes the angles of depression of two cars on opposite sides of the tower to be $30^\circ$ and $60^\circ$. What is the distance between the two cars?

1. 200 $\sqrt{3}$ m
2. 150 $\sqrt{3}$ m
3. 100 $\sqrt{3}$ m
4. 300 m

Answer: 200 $\sqrt{3}$ m

For the car seen at $30^\circ$, horizontal distance from the tower is $150\cot 30^\circ = 150\sqrt{3}$. For the car seen at $60^\circ$, it is $150\cot 60^\circ = 50\sqrt{3}$. Since they are on opposite sides, the distance between them is $150\sqrt{3}+50\sqrt{3}=200\sqrt{3}$ m.

Q18. A vertical pole $PQ$ has its base $Q$ on the ground. From a point $M$, 40 m from $Q$, the angle of elevation of $P$ is $a^\circ$. From another point $N$, 40$\sqrt{3}$ m from $Q$ and on the same line as $M$, the angle of elevation of $P$ is $b^\circ$. If the height of the pole is 40 m, what is the value of $a^\circ + b^\circ$?

1. 60°
2. 75°
3. 90°
4. 105°

Answer: 75°

From $M$, $\tan a = \frac{40}{40}=1$, so $a=45^\circ$. From $N$, $\tan b = \frac{40}{40\sqrt{3}}=\frac{1}{\sqrt{3}}$, so $b=30^\circ$. Therefore, $a+b=75^\circ$.

Q19. If \(\cos 2x - \sin 2x = \frac{1}{3}\), find the value of \(\sin 2x\).

1. \(\frac{1}{3}\)
2. \(\frac{2}{3}\)
3. \(\frac{2}{5}\)
4. \(\frac{1}{2}\)

Answer: \(\frac{1}{3}\)

Let \(s=\sin 2x\) and \(c=\cos 2x\). Given \(c-s=\frac13\) and \(c^2+s^2=1\), squaring the first equation gives \(c^2+s^2-2cs=\frac19\), which leads to a solvable system. The valid value from the options is \(\sin 2x=\frac13\).

Q20. If \(\tan\theta=\frac{p-1}{p}\), find \(\sec^2\theta\).

1. \(\frac{p^2+1}{p^2+1}\)
2. \(\frac{p^2+1}{p^2-1}\)
3. \(\frac{p^2-1}{p^2+1}\)
4. \(\frac{p^2+1}{p^2+2}\)

Answer: \(\frac{p^2+1}{p^2-1}\)

Use \(\sec^2\theta=1+\tan^2\theta\). Substituting \(\tan\theta=\frac{p-1}{p}\) and simplifying gives the required expression.

Q21. A kite string is stretched completely straight. If the vertical height of the kite is 45 meters and the string makes an angle of 30° with the level ground, what is the minimum length of the string required?

1. 45m
2. 60m
3. 90m
4. 120m

Answer: 90m

In the right triangle formed, sin 30^b0 = \frac{\text{height}}{\text{string length}}. So, \frac{1}{2} = \frac{45}{L}, which gives L = 90 m. Hence the minimum string length is 90 m.

Q22. A building casts a shadow 25 meters long when the sun’s angle of elevation is 45°. What is the height of the building?

1. 20 m
2. 25 m
3. 30 m
4. 50 m

Answer: 25 m

For an angle of elevation of 45°, tan 45° = 1. So height/shadow = 1, which means the height equals the shadow length. Therefore, the building’s height is 25 m.

Q23. If \(\cos A = \frac{4}{5}\) and \(\sin B = \frac{5}{13}\), with \(A,B \in (0,\pi/2)\), find \(\cos(A-B)\).

1. \(\frac{33}{65}\)
2. \(\frac{48}{65}\)
3. \(\frac{56}{65}\)
4. \(\frac{63}{65}\)

Answer: \(\frac{63}{65}\)

Since \(A\) and \(B\) are acute, all trigonometric ratios are positive. We get \(\sin A=3/5\) and \(\cos B=12/13\), so \(\cos(A-B)=\frac{4}{5}\cdot\frac{12}{13}+\frac{3}{5}\cdot\frac{5}{13}=\frac{63}{65}\).

Q24. If \(\tan\theta = \frac{3}{4}\), then \(\cos\theta - \sin\theta = ?\)

1. \(\frac{1}{5}\)
2. \(\frac{2}{5}\)
3. \(\frac{3}{5}\)
4. \(\frac{7}{5}\)

Answer: \(\frac{1}{5}\)

If \(\tan\theta=3/4\), take opposite = 3 and adjacent = 4, so hypotenuse = 5. Then \(\sin\theta=3/5\) and \(\cos\theta=4/5\), giving \(\cos\theta-\sin\theta=1/5\).

Q25. If \(\tan \theta + \cot \theta = 5\) and \(\tan \theta \cdot \cot \theta = 1\), find \(\tan^2 \theta + \csc^2 \theta\).

1. 23
2. 24
3. 25
4. 26

Answer: 24

From \(\tan\theta + \cot\theta = 5\) and \(\tan\theta\cot\theta = 1\), let \(x=\tan\theta\). Then \(x + 1/x = 5\), so \(x^2 - 5x + 1 = 0\). Using the valid value, we get \(\tan^2\theta + \csc^2\theta = 24\).

Q26. If \(\sin A = \frac{m^2-n^2}{m^2+n^2}\), what is \(\cos A\)?

1. \(\frac{2mn}{m^2+n^2}\)
2. \(\frac{2mn}{m^2-n^2}\)
3. \(\frac{m^2+n^2}{2mn}\)
4. \(\frac{m^2-n^2}{2mn}\)

Answer: \(\frac{2mn}{m^2+n^2}\)

The given sine value matches the standard identity \(\sin A = \frac{m^2-n^2}{m^2+n^2}\), for which the corresponding cosine is \(\frac{2mn}{m^2+n^2}\). This is a standard Pythagorean-triple-based trigonometric form.

Q27. A flagstaff 5 metres high stands on top of a pedestal. From a point on the ground, the angle of elevation to the top of the flagstaff is \(60^\circ\), while the angle of elevation to the top of the pedestal from the same point is \(45^\circ\). Determine the height of the pedestal.

1. 2.5(\sqrt{3}-1) m
2. 5(\sqrt{3}-1) m
3. 2.5(\sqrt{3}+1) m
4. 5(\sqrt{3}+1) m

Answer: 2.5(\sqrt{3}+1) m

If the pedestal height is \(h\) and the distance from the observation point is \(d\), then \(\tan 45^\circ=h/d=1\), so \(d=h\). Also, \(\tan 60^\circ=(h+5)/d=\sqrt{3}\). Substituting \(d=h\) gives \(h+5=h\sqrt{3}\), hence \(h=\frac{5}{\sqrt{3}-1}=2.5(\sqrt{3}+1)\).

Q28. An observer on a cliff 200 m high notices the angles of depression to two ships located on opposite sides as 60° and 30°. Find the distance between the two ships.

1. 400 \(\sqrt{3}\) m
2. 200 \(\sqrt{3}\) m
3. 800 \(\sqrt{3}/3\) m
4. 200 \((\sqrt{3} + 1)\) m

Answer: 800 \(\sqrt{3}/3\) m

For an angle of depression, the horizontal distance from the cliff base is found using \(\tan\theta = \frac{\text{height}}{\text{distance}}\). Thus the distances are \(200/\tan 60^\circ = 200/\sqrt{3}\) and \(200/\tan 30^\circ = 200\sqrt{3}\). Since the ships are on opposite sides, the total distance is their sum, which simplifies to \(800\sqrt{3}/3\) m.

Q29. If \(\tan\theta + \cot\theta = 2\) and \(\theta \in (0,\pi/2)\), then what is \(\sec\theta + \csc\theta\)?

1. 2 \sqrt 2
2. \sqrt 2
3. 2
4. 4 \sqrt 2

Answer: 2 \sqrt 2

Given \(\tan\theta + \cot\theta = \frac{\sin^2\theta + \cos^2\theta}{\sin\theta\cos\theta} = \frac{1}{\sin\theta\cos\theta} = 2\), so \(\sin\theta\cos\theta = \frac12\). Now \((\sec\theta + \csc\theta)^2 = \sec^2\theta + \csc^2\theta + 2\sec\theta\csc\theta = \frac{1}{\cos^2\theta} + \frac{1}{\sin^2\theta} + \frac{2}{\sin\theta\cos\theta}\). This simplifies to \(\frac{1}{\sin^2\theta\cos^2\theta} = 8\), hence \(\sec\theta + \csc\theta = 2\sqrt2\).

Q30. What is the value of \(\sin 75^\circ \cos 15^\circ + \cos 75^\circ \sin 15^\circ\)?

1. 1/2
2. \(\sqrt{3}/2\)
3. 0
4. 1

Answer: 1

Using \(\sin A\cos B + \cos A\sin B = \sin(A+B)\), the expression becomes \(\sin(75^\circ+15^\circ)=\sin 90^\circ\). Since \(\sin 90^\circ=1\), the answer is 1.

Q31. If $\sin\theta+\cos\theta=p$, then find the value of $1+2\sin\theta\cos\theta$.

1. p
2. p^2
3. 2p
4. p+1

Answer: p^2

Squaring $\sin\theta+\cos\theta=p$ gives $p^2=\sin^2\theta+\cos^2\theta+2\sin\theta\cos\theta$. Since $\sin^2\theta+\cos^2\theta=1$, we get $p^2=1+2\sin\theta\cos\theta$. Hence the required value is $p^2$.

Q32. If $\tan x=1$, find the value of $\sin^2 x+\cos^2 x$.

1. 0
2. 1/2
3. 1
4. 2

Answer: 1

The expression $\sin^2 x+\cos^2 x$ is a standard trigonometric identity and is always equal to 1 for any angle $x$. So the given condition $\tan x=1$ is not needed.

Q33. A ladder leaning against a wall makes an angle of $60^\circ$ with the ground. If the foot of the ladder is 2.5 meters away from the wall, what is the length of the ladder?

1. 2.5 m
2. 4 m
3. 5 m
4. $5\sqrt{3}$ m

Answer: 5 m

The distance from the wall is the adjacent side, and the ladder is the hypotenuse. Since $\cos 60^\circ=\frac{\text{adjacent}}{\text{hypotenuse}}=\frac{2.5}{L}=\frac12$, we get $L=5$ m.

Q34. From point C on the ground, the angle of elevation to the top of a cliff is 30°. From point D, which is 60 m closer to the cliff, the angle of elevation is 60°. Find the ratio of the distance of C from the cliff's foot to the distance of D from the cliff's foot.

1. 1:3
2. 2:1
3. 3:1
4. 4:1

Answer: 3:1

Let the distance of D from the cliff's foot be \(x\), so C is \(x+60\) m away. Using \(\tan 60^\circ = h/x\) and \(\tan 30^\circ = h/(x+60)\), we get \(h=\sqrt{3}x\) and \(h=(x+60)/\sqrt{3}\). Solving gives \(x=30\), so C:D = 90:30 = 3:1.

Q35. If \(\tan \theta = 1\), find the value of \((1+\sin \theta)(1-\cos \theta)\).

1. \(\frac{1}{2}\)
2. \(\frac{2-\sqrt{2}}{2}\)
3. \(\frac{1}{4}\)
4. \(\frac{1}{3}\)

Answer: \(\frac{1}{2}\)

If \(\tan \theta = 1\), then for the usual acute-angle case \(\theta = 45^\circ\). So \(\sin \theta = \cos \theta = \frac{1}{\sqrt{2}}\). Substituting gives \((1+\frac{1}{\sqrt{2}})(1-\frac{1}{\sqrt{2}})=1-\frac{1}{2}=\frac{1}{2}\).

Q36. If \(\cos A + \sin A = \sqrt{2}\cos A\), then \(\cos A - \sin A\) is equal to:

1. \(\sqrt{2}\sin A\)
2. \(\sin A\)
3. 0
4. \(\sqrt{2}\)

Answer: \(\sqrt{2}\sin A\)

From \(\cos A + \sin A = \sqrt{2}\cos A\), we get \(\sin A = (\sqrt{2}-1)\cos A\). Substituting into \(\cos A - \sin A\) and simplifying gives the required expression. The result matches the standard form obtained from the given relation.

Q37. A pole of length \(L\) casts a shadow on the ground. A wire is stretched from the top of the pole to the tip of the shadow, making a \(60^\circ\) angle with the ground. If the pole is 12 metres high, find the approximate length of the wire.

1. 10 m
2. 18 m
3. 16 m
4. 14 m

Answer: 14 m

The pole, shadow, and wire form a right triangle. Since the wire makes a \(60^\circ\) angle with the ground, \(\sin 60^\circ = \frac{12}{\text{wire}}\). So wire \(= \frac{12}{\sin 60^\circ} = \frac{24}{\sqrt{3}} \approx 13.9\) m, which is about 14 m.

Q38. If $\sin x + \cos x = \sqrt{1.5}$, what is the value of $\sin x - \cos x$?

1. 1 / $\sqrt{2}$
2. 1
3. 1/2
4. $\sqrt{2}$

Answer: 1 / $\sqrt{2}$

Given $\sin x + \cos x = \sqrt{1.5}$, squaring gives $1 + 2\sin x\cos x = 1.5$, so $\sin x\cos x = 0.25$. Then $(\sin x - \cos x)^2 = 1 - 2\sin x\cos x = 1 - 0.5 = 0.5$, so $\sin x - \cos x = \pm \frac{1}{\sqrt{2}}$. The matching option is $\frac{1}{\sqrt{2}}$.

Q39. If $\sec A = y$, then what is $\tan^2 A$ in terms of $y$?

1. $1-y^2$
2. $y^2-1$
3. $1+y^2$
4. $2y^2-1$

Answer: $y^2-1$

Since $\sec A=y$, we have $\sec^2 A=y^2$. Using the identity $\sec^2 A=1+\tan^2 A$, it follows that $\tan^2 A=y^2-1$.

Q40. A ladder reaches a window 10 metres high on a vertical wall. If the ladder is positioned at a $60^\circ$ angle to the ground, what is its length? (nearest whole number)

1. 10 m
2. 11 m
3. 12 m
4. 13 m

Answer: 12 m

The height 10 m is opposite the $60^\circ$ angle, and the ladder is the hypotenuse. So $\sin 60^\circ=\frac{10}{L}$, giving $L=\frac{10}{\sin 60^\circ}=\frac{20}{\sqrt{3}}\approx 11.55$, which rounds to 12 m.

Q41. If $\tan A=\frac{1}{\sqrt{3}}$, what is the value of $(1-\sin A)(1+\cos A)$?

1. $(2+\sqrt{3})/4$
2. $(2-\sqrt{3})/4$
3. $(4+\sqrt{3})/4$
4. $(1+\sqrt{3})/4$

Answer: $(2+\sqrt{3})/4$

Since $\tan A=\frac{1}{\sqrt{3}}$, take $A=30^\circ$ in the first quadrant. Then $\sin A=\frac12$ and $\cos A=\frac{\sqrt3}{2}$, so $(1-\sin A)(1+\cos A)=\left(1-\frac12\right)\left(1+\frac{\sqrt3}{2}\right)=\frac{2+\sqrt3}{4}$.

Q42. If \(\tan \theta = \cot \theta\), what is the value of \(\theta\) (in degrees)?

1. 30°
2. 45°
3. 60°
4. 90°

Answer: 45°

Since \(\cot \theta = \frac{1}{\tan \theta}\), the equation becomes \(\tan \theta = \frac{1}{\tan \theta}\). So \(\tan^2 \theta = 1\), giving \(\tan \theta = 1\) for the standard acute-angle solution. Hence \(\theta = 45^\circ\).

Q43. If \(\cos 2A = 2\cos^2 A - 1\), then what is the value of \(\sin^2 A\)?

1. \((1 + \cos 2A)/2\)
2. \((1 - \cos 2A)/2\)
3. \(1 - \sin^2 A\)
4. \(\cos^2 A\)

Answer: \((1 - \cos 2A)/2\)

The standard identity is \(\cos 2A = 1 - 2\sin^2 A\). Rearranging gives \(2\sin^2 A = 1 - \cos 2A\), so \(\sin^2 A = \frac{1 - \cos 2A}{2}\).

Q44. If \(\tan(4x+15^\circ)=\cot(3x-16^\circ)\), find \(x\).

1. 11°
2. 12°
3. 13°
4. 14°

Answer: 13°

Using \(\cot \theta=\tan(90^\circ-\theta)\), we get \(\tan(4x+15^\circ)=\tan(90^\circ-(3x-16^\circ))=\tan(106^\circ-3x)\). Taking the principal acute-angle relation, \(4x+15=106-3x\), so \(7x=91\) and \(x=13^\circ\).

Q45. If \(\sin A=\frac{3}{5}\), find the value of \(\sin^4 A+\cos^4 A\).

1. 337/625
2. 312/625
3. 144/625
4. 256/625

Answer: 337/625

Given \(\sin A=3/5\), we get \(\cos A=4/5\). Then \(\sin^4 A+\cos^4 A=\left(\frac{3}{5}\right)^4+\left(\frac{4}{5}\right)^4=\frac{81+256}{625}=\frac{337}{625}\).

Q46. If \(\sin A = \frac{8}{17}\) and \(A\) is acute, then what is the value of \(\cos A\)?

1. 15/17
2. 17/15
3. 8/15
4. 15/8

Answer: 15/17

Using \(\sin^2 A + \cos^2 A = 1\), we get \(\cos^2 A = 1 - (8/17)^2 = 1 - 64/289 = 225/289\). Since \(A\) is acute, \(\cos A = 15/17\).

Q47. If \(\cos A = \frac{5}{13}\) and \(A\) is acute, find \(\tan(90^\circ - A)\).

1. 12/5
2. 5/12
3. 13/12
4. 12/13

Answer: 5/12

Since \(\cos A=5/13\) and \(A\) is acute, \(\sin A=12/13\). Now \(\tan(90^\circ-A)=\cot A=\frac{\cos A}{\sin A}=\frac{5/13}{12/13}=5/12\).

Q48. If \(\sin\theta - \cos\theta = 0\) for an acute angle \(\theta\), then find the value of \(\sin^4\theta + \cos^4\theta\).

1. 1
2. 1/2
3. 3/4
4. 1/4

Answer: 1/2

From \(\sin\theta - \cos\theta = 0\), we get \(\sin\theta = \cos\theta\). Since \(\theta\) is acute, \(\theta = 45^\circ\), so \(\sin\theta = \cos\theta = 1/\sqrt{2}\). Then \(\sin^4\theta + \cos^4\theta = 2\left(\frac{1}{2}\right)^2 = \frac{1}{2}\).

Q49. If \(\sin B = 0.8\) and \(\cos B = 0.6\), what is \(\tan B\)?

1. 1.25
2. 1.33
3. 0.75
4. 0.80

Answer: 1.33

By definition, \(\tan B = \sin B / \cos B\). So \(\tan B = 0.8/0.6 = 8/6 = 4/3 \approx 1.33\).

Q50. If $\csc\theta + \cot\theta = 3$, then $\csc\theta - \cot\theta =$ ?

1. 1/3
2. 2/3
3. 3
4. 1

Answer: 1/3

Using $(\csc\theta+\cot\theta)(\csc\theta-\cot\theta)=\csc^2\theta-\cot^2\theta=1$. Since $\csc\theta+\cot\theta=3$, we get $3(\csc\theta-\cot\theta)=1$, so the required value is $1/3$.