$\frac{\cos 28^{\circ}}{\sin 62^{\circ}}=?$
- A$2$
- B$-1$
- C$0$
- ✓$1$
Answer: D.
View full solution →Question types
Trigonometry question types
42 questions across 6 question groups — pick any mix to generate a Maths paper with step-by-step answer keys.
42
Questions
6
Question groups
5
Question types
01
MCQ(1M)
4 Q→021 Mark Question
3 Q→032 Mark Question
25 Q→043 Mark Question
1 Q→054 Mark Question
4 Q→065 Mark Question
5 Q→Sample Questions
Trigonometry questions
One sample from each question group in this chapter. Select any group above to see the full set with answer keys.
2 tan 45° + cos 45° – sin 45° = ?
- A0
- B1
- ✓2
- D3
Answer: C.
View full solution →Which of the following is the value of sin 90°?
- A$\frac{\sqrt{3}}{2}$
- B0
- C$\frac{1}{2}$
- ✓1
Answer: D.
View full solution →Which of the following statements is true?
- ✓sin θ = cos (90 – θ)
- Bcos θ = tan (90 – θ)
- Csin θ = tan (90 – θ)
- Dtan θ = tan (90 – θ)
Answer: A.
View full solution →In right angled $\triangle \mathrm{ACB}$, If $\angle \mathrm{C}=90^{\circ}, \mathrm{AC}=3, \mathrm{BC}=4$.Find the ratios $\sin A, \sin B, \cos A, \tan B$
View full solution →Find the value of $\frac{\cos 56^{\circ}}{\sin 34^{\circ}}$
View full solution →Find the value of 2tan 45° + cos 30° - sin 60°
Conduct the above discussed activity and find the height of a tall tree in your surrounding. If there is no tree in the premises, then find the height of a pole.
It is convenient to do the above experiment between 11:30 am and 1:30 pm instead of doing it in the morning at 8’O clock. Can you tell why?
Measuring height of a tree using trigonometric ratios.
This experiment can be conducted on a clear sunny day. Look at the figure given above. Height of the tree is QR, height of the stick is BC.
Thrust a stick in the ground as shown in the figure. Measure its height and length of its shadow. Also measure the length of the shadow of the tree. Using these values, how will you determine the height of the tree?
While solving the above illustrative example, can we take the lengths of PQ and PR as 5 and 13? If so, then what changes are needed In the writing of the solution.
While solving the above Illustrative example, why the lengths of PQ and PR are taken 5k and 13k?
In right angled triangle $\triangle \mathrm{PQR}, \angle \mathrm{Q}=90^{\circ}, \angle \mathrm{R}=\theta$ and if $\sin \theta=\frac{5}{13}$ then find $\cos \theta$ and $\tan \theta$
View full solution →Verify that the equation $‘sin2 \theta + cos2 \theta = 1’$ is true when $\theta = 0^\circ$ or $\theta = 90^\circ.$
View full solution →In right angled $\triangle P Q R, \angle Q=90^{\circ}, \angle R=\theta$ and if $\sin \theta=\frac{5}{13}$, then find $\cos \theta$ and $\tan \theta$,
View full solution →If $\cos \theta=\frac{15}{17}$, then find $\sin \theta$.
View full solution →If $\sin \theta=\frac{4}{5}$, then find $\cos \theta$.
View full solution →Fill in the blanks.
In right angled $\triangle LMN$, if $\angle N =\theta, \angle M =90^{\circ}, \cos \theta=\frac{24}{25}$, find $\sin \theta$ and $\tan \theta$. Similarly, find $\left(\sin ^2 \theta\right)$ and $\left(\cos ^2 \theta\right)$
View full solution →In right angled $\triangle Y X Z, \angle X=90^{\circ}, X Z=8 cm, Y Z=17 cm$, find $\sin Y, \cos Y, \tan Y, \sin Z, \cos Z, \tan Z$.
View full solution →In right angled $\triangle T S U, T S=5, \angle S=90^{\circ}, S U=12$, then find $\sin T, \cos T$, $\tan T$. Similarly find $\sin U, \cos U, \tan U$.
View full solution →In right angled ∆PQR, ∠Q = 900. Therefore ∠P and ∠R are complementary angles of each other. Verify the following ratios.
i. sin θ = cos (90 – θ)
ii. cos θ = sin (90 – θ)
iii. sin 30° = cos (90° – 30°) = cos 60°
iv. cos 30° = sin (90° – 30°) = sin 60°
i. sin θ = cos (90 – θ)
ii. cos θ = sin (90 – θ)
iii. sin 30° = cos (90° – 30°) = cos 60°
iv. cos 30° = sin (90° – 30°) = sin 60°
Generate a Trigonometry paper free
Pick question groups from the list above, set marks and difficulty, and export a branded PDF with step-by-step answer keys. First 3 chapters free — no signup.