Chapter 8 Introduction to Trigonometry- MCQ Online Test 3 Class 10 Maths
1. Find the value of sin 60°cos 30° + sin 30°cos 60°?
a. 2
b. 3
c. 5
d. 1
2. Find the value of 2 tan²45°+ cos²30°−sin²60°?
a. 1
b. 2
c. 5
d. 3
3. If sin θ − cos θ = 0, then find the value of θ?
a. 35∘
b. 45∘
c. 55∘
d. 25∘
4. Find the value of tan1∘ tan 2∘tan 3∘………… tan 89∘
a. 5
b. 3
c. 4
d. 1
5. If ΔABC is right angled at C, then find the value of cos (A + B)?
a. 1
b. 5
c. 0
d. 3
6. Find the value of cos 48°−sin 42°?
a. 0
b. 3
c. 5
d. 2
7. If A and B are acute angles and tan A = cot B, then find the value of (A + B)?
a. 30∘
b. 20∘
c. 90∘
d. 60∘
8. If tan 2A = cot (A−18°), then find the value of ‘A’?
a. 15°
b. 36°
c. 23°
d. 43°
9. If ΔPQR is right angled at Q, then find the value of sin (P + R)?
a. 3
b. 2
c. 1
d. 8
10. If A and B are acute angles and sin A = cos B, then find the value of (A + B)?
a. 10°
b. 30°
c. 60°
d. 90°
Chapter - 8 Introduction to Trigonometry Quiz-2 | Math Class 10th
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Question 1 of 10
1. Question
Find the value of sin60°cos30° + sin30°cos60°?
Correct
Incorrect
-
Question 2 of 10
2. Question
Find the value of 2 tan²45°+ cos²30°−sin²60°?
Correct
Incorrect
-
Question 3 of 10
3. Question
If sin θ − cos θ = 0, then find the value of θ?
Correct
Given: sinθ – cosθ = 0
sinθ = cosθ
sinθ = sin (90∘−θ)
θ = 90∘−θ
2θ = 90∘
θ = 45∘Incorrect
Given: sinθ – cosθ = 0
sinθ = cosθ
sinθ = sin (90∘−θ)
θ = 90∘−θ
2θ = 90∘
θ = 45∘ -
Question 4 of 10
4. Question
Find the value of tan 1∘ tan 2∘tan 3∘………… tan 89∘
Correct
tan1°tan 2°tan 3°………… tan 89°
= tan(90°−89°)tan(90°−88°)tan(90°−87°)……….tan87°tan88°tan89°
= cot89°cot88°cot87°……….tan87°tan88°tan89°
= (cot89°tan89°)(cot88°tan88°)(cot87°tan87°)…..(cot44°tan44°)tan45°= 1 x 1 x 1 x 1 x 1 …………….. 1 = 1
Incorrect
tan1°tan 2°tan 3°………… tan 89°
= tan(90°−89°)tan(90°−88°)tan(90°−87°)……….tan87°tan88°tan89°
= cot89°cot88°cot87°……….tan87°tan88°tan89°
= (cot89°tan89°)(cot88°tan88°)(cot87°tan87°)…..(cot44°tan44°)tan45°= 1 x 1 x 1 x 1 x 1 …………….. 1 = 1
-
Question 5 of 10
5. Question
If ΔABC is right angled at C, then find the value of cos (A + B)?
Correct
Let ∠A = θ If in a right angled triangle ABC, right angled at C, then ∠B = 90°− θ
Now, cos (A + B) = cos (θ+90°−θ)
= cos 90°
= 0Incorrect
Let ∠A = θ If in a right angled triangle ABC, right angled at C, then ∠B = 90°− θ
Now, cos (A + B) = cos (θ+90°−θ)
= cos 90°
= 0 -
Question 6 of 10
6. Question
Find the value of cos 48°−sin 42°?
Correct
Here, cos 48°−sin 42°
= cos 48° − sin (90°−48°)
= cos 48° − cos 48° = 0Incorrect
Here, cos 48°−sin 42°
= cos 48° − sin (90°−48°)
= cos 48° − cos 48° = 0 -
Question 7 of 10
7. Question
If A and B are acute angles and tan A = cot B, then find the value of (A + B)?
Correct
Given: tan A = cot B
since A and B are acute angles, then
tan A = tan (90°−B)
A = 90°− B
A + B = 90°Incorrect
Given: tan A = cot B
since A and B are acute angles, then
tan A = tan (90°−B)
A = 90°− B
A + B = 90° -
Question 8 of 10
8. Question
If tan 2A = cot (A−18°), then find the value of ‘A’?
Correct
Given: tan 2A = cot (A−18°)
cot (90°−2A) = cot (A−18°)
2A + A = 90° + 18°
3A = 108°
A = 36°Incorrect
Given: tan 2A = cot (A−18°)
cot (90°−2A) = cot (A−18°)
2A + A = 90° + 18°
3A = 108°
A = 36° -
Question 9 of 10
9. Question
If ΔPQR is right angled at Q, then find the value of sin (P + R)?
Correct
If in right angled triangle PQR, right angled at Q, then P and R are acute angles.
Let ∠P = θ, then ∠R
= 90−θ
Now, sin (P + R) = sin (θ+90−θ)= sin90
= 1
Incorrect
If in right angled triangle PQR, right angled at Q, then P and R are acute angles.
Let ∠P = θ, then ∠R
= 90−θ
Now, sin (P + R) = sin (θ+90−θ)= sin90
= 1
-
Question 10 of 10
10. Question
If A and B are acute angles and sin A = cos B, then find the value of (A + B)?
Correct
Given: sin A = cos B
Since A and B are acute angles, then
sin A = sin (90°−B)
A = 90°−B
A + B = 90°Incorrect
Given: sin A = cos B
Since A and B are acute angles, then
sin A = sin (90°−B)
A = 90°−B
A + B = 90°