Take sectional tests or a full 40-question Trigonometry & Co-ordinate Geometry mock for CUET UG with timers, answer review, and detailed explanations.
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Use the sectional sessions to isolate standard-angle recall, heights and distances, coordinate formulas, and mixed slope-collinearity questions. Then switch to the full mixed mock to test speed under timer pressure.
One question at a time, 60 seconds per question, with score, accuracy, and subtopic insight at the end.
1. What is the value of sin²θ + cos²θ?
Explanation: This is the basic Pythagorean identity.For every angle, $\sin^2\theta + \cos^2\theta = 1$.
2. What is tan 45deg?
Explanation: $\tan 45^\circ = 1$ because $\sin 45^\circ = \cos 45^\circ$.
3. What is sin 30deg + cos 60deg?
Explanation: $\sin 30^\circ = 1/2$ and $\cos 60^\circ = 1/2$.So the sum is 1.
4. What is sec²θ - tan²θ equal to?
Explanation: From $1+\tan^2\theta=\sec^2\theta$, we get $\sec^2\theta-\tan^2\theta=1$.
5. What is cos 0deg × sin 90deg + sin 0deg × cos 90deg?
Explanation: $\cos 0^\circ=1$, $\sin 90^\circ=1$, $\sin 0^\circ=0$, and $\cos 90^\circ=0$.So the value is $1\times1 + 0\times0 = 1$.
6. Find the value of 4cos²60deg - sin⁴30deg.
Explanation: $4\times(1/2)^2 - (1/2)^4 = 4\times1/4 - 1/16 = 1 - 1/16 = 15/16$.
7. If sinθ = 5/8, what is cosθ for an acute angle?
Explanation: Take perpendicular = 5 and hypotenuse = 8.Then base $=\sqrt{8^2-5^2}=\sqrt{39}$, so $\cos\theta=\sqrt{39}/8$.
8. What is tan 860deg equal to?
Explanation: $860^\circ = 180^\circ\times 5 - 40^\circ$.Using $\tan(180n-\theta)=-\tan\theta$, the value is $-\tan 40^\circ$.
9. 485deg lies in which quadrant?
Explanation: $485^\circ = 360^\circ + 125^\circ$.Since $125^\circ$ lies between $90^\circ$ and $180^\circ$, it is in the second quadrant.
10. Find sin²45deg + sin²30deg + sin²60deg + sin²90deg.
Explanation: $1/2 + 1/4 + 3/4 + 1 = 2.5$.
11. From the top of a 100 m tower, the angle of depression of a car is 30deg. Find the horizontal distance of the car.
Explanation: Angle of depression equals angle of elevation.So $\tan 30^\circ = 100/x$, giving $x = 100\sqrt{3}$ m.
12. A kite is flying with a 150 m thread making a 30deg angle with the horizontal. Find the height of the kite.
Explanation: Thread is the hypotenuse.Height $=150\sin 30^\circ = 150\times1/2 = 75$ m.
13. A 200 m tower has shadow lengths corresponding to sun elevations 45deg and 60deg. By how much does the shadow shorten?
Explanation: At $45^\circ$, shadow = 200 m.At $60^\circ$, shadow = $200/\sqrt{3}$.Difference = $200 - 200/\sqrt{3} = 200(3-\sqrt{3})/3$.
14. From a point 20 m from the foot of a tower, the angle of elevation is 30deg. Find the height of the tower.
Explanation: $\tan 30^\circ = h/20$.So $h = 20/\sqrt{3}$ m.
15. Two trees are 10 m and 18 m high. The distance between their tops is 17 m. Find the horizontal distance between the trees.
Explanation: The vertical difference is 8 m.By Pythagoras, horizontal distance $=\sqrt{17^2-8^2}=15$ m.
16. A tree breaks and its top touches the ground making a 30deg angle at a point 50 m from the foot. Find the original height.
Explanation: Remaining vertical part $=50/\sqrt{3}$ and broken part $=100/\sqrt{3}$.Total original height $=150/\sqrt{3}$ m.
17. From two points 2 km apart on the same straight line, the angles of elevation of an aeroplane are 45deg and 60deg. Find the height of the aeroplane above the ground.
Explanation: Let the nearer distance be $x$ km.Then $h=x\sqrt{3}$ and also $h=x+2$.Solving gives $x=\sqrt{3}+1$ and $h=x\sqrt{3}=3+\sqrt{3}$ km.
18. From a cliff 100 m high, the angles of depression of the top and bottom of a tower are 30deg and 60deg. Find the height of the tower.
Explanation: If horizontal distance is $d$, then $\tan 60^\circ = 100/d$ so $d = 100/\sqrt{3}$.Also $\tan 30^\circ = (100-h)/d$.So $(100-h)=100/3$, which gives $h=200/3$ m.
19. The shadow of a tree at 45deg sun elevation is 20 m longer than its shadow at 60deg. Find the height of the tree.
Explanation: Let height be $h$.Then shadows are $h$ and $h/\sqrt{3}$.So $h-h/\sqrt{3}=20$, giving $h=10\sqrt{3}(\sqrt{3}+1)\approx47.32$ m.
20. The angle of elevation of a tower from 60 m away is 30deg. From 20 m away, what is the angle of elevation?
Explanation: Height of the tower = $60\tan 30^\circ = 20\sqrt{3}$.At 20 m, $\tan\theta = 20\sqrt{3}/20 = \sqrt{3}$, so $\theta=60^\circ$.
21. Find the distance between (3,4) and (0,0).
Explanation: $d=\sqrt{3^2+4^2}=\sqrt{25}=5$.
22. Find the midpoint of the segment joining (4,-2) and (-2,6).
Explanation: Midpoint $=\left((4-2)/2,(-2+6)/2\right)=(1,2)$.
23. What is the slope of the line joining (2,3) and (5,9)?
Explanation: $m=(9-3)/(5-2)=6/3=2$.
24. What is the slope of a line perpendicular to a line with slope 3?
Explanation: For perpendicular lines, $m_1m_2=-1$.So the required slope is $-1/3$.
25. In what ratio does the x-axis divide the segment joining (4,-6) and (1,3)?
Explanation: At the x-axis, $y=0$.Using section formula for y-coordinate: $(3k-6)/(k+1)=0$ gives $k=2$.
26. Find the centroid of the triangle with vertices (2,4), (6,2), and (4,6).
Explanation: Centroid $=\left((2+6+4)/3,(4+2+6)/3\right)=(4,4)$.
27. Two vertices of a triangle are (-2,5) and (-4,4), and its centroid is at the origin. Find the third vertex.
Explanation: If the centroid is $(0,0)$, then the sum of x-coordinates and the sum of y-coordinates are both 0.So $x_3=6$ and $y_3=-9$.
28. The points (2,-k), (0,-5), and (5/2,0) are collinear. Find k.
Explanation: Using the area-zero condition for collinear points gives $k=1$.
29. What is the slope of the line joining (2,6) and (-3,1)?
Explanation: $m=(6-1)/(2-(-3))=5/5=1$.
30. A point equidistant from (3,0) and (-3,0) must lie on which line?
Explanation: The perpendicular bisector of the segment joining $(3,0)$ and $(-3,0)$ is the y-axis.
31. If the distance between (x,6) and (3,0) is 10, what are the possible values of x?
Explanation: $\sqrt{(x-3)^2+36}=10$ gives $(x-3)^2=64$.So $x=11$ or $x=-5$.
32. A point (x,y) is equidistant from (a+b, b-a) and (a-b, a+b). Which relation follows?
Explanation: Equating squared distances and simplifying gives $bx = ay$.
33. If the slope of one line is 1, then the slope of a perpendicular line is:
Explanation: For perpendicular lines, product of slopes is -1.So the perpendicular slope to 1 is -1.
34. Two non-vertical parallel lines have slopes:
Explanation: Parallel lines run in the same direction, so their slopes are equal.
35. What is the distance of the point (6,8) from the origin?
Explanation: Distance from origin $=\sqrt{6^2+8^2}=\sqrt{100}=10$.
36. Which point is the midpoint of (1,1) and (5,7)?
Explanation: Midpoint $=\left((1+5)/2,(1+7)/2\right)=(3,4)$.
37. Which of the following equals cos 30deg?
Explanation: $\cos 30^\circ = \sqrt{3}/2$.
38. In which quadrant are tan and cot both positive?
Explanation: Tan and cot are positive wherever sin and cos have the same sign: Quadrants I and III.
39. If A=(0,0) and B=(8,15), then AB is:
Explanation: $AB=\sqrt{8^2+15^2}=\sqrt{289}=17$.
40. Which formula is used to test whether three coordinate points are collinear?
Explanation: Three points are collinear if the area of the triangle formed by them is zero.