Ncert – Class 10 – Introduction to Trigonometry – Hots Question

Question 1

If triangle ABC is right angled at C, in which AB = 50 units, BC = 48 units and ∠ABC =α
Determine the values of cosα + sinα?

Solution

$$ cos\alpha + sin \alpha = (\frac{14}{50}+\frac{48}{50})=\frac{62}{50}$$

Question 2

Prove that sec⁴A(1-sin⁴A) – 2 tan²A =1

Solution

LHS,
sec⁴A(1-sin⁴A) – 2 tan2A
sec⁴A – sec⁴A sin⁴A- 2 tan²A
sec⁴A – (𝑠𝑖𝑛𝐴/𝑐𝑜𝑠𝐴)⁴ – 2 tan²A { 𝑠𝑖𝑛𝐴/𝑐𝑜𝑠𝐴= tanA}
sec⁴A – tan⁴A – 2 tan²A
(sec²A)² – (tan²A)² – 2 tan²A
sec2A + tan2A – 2 tan2A
sec2A – tan2A = 1 = RHS

Question 3

In the diagram below, BC is perpendicular to AD, and BD is 10m, ∠ACB = 45° and ∠BCD = 30°, Find AB.

Solution

10√3

Question 4

If sin (A – B) = ½ and cos (A + B) = ½, where (A + B) ≤ 900² and A > B.
1 (a) Find the values of A and B.
1 (b) Find the value of tan 2A

Solution

1(a) A = 45º; B = 15º
Sin (A – B) = ½
⇒ sin (A – B) = sin 30º
⇒ A– B = 30º——(i)
Cos (A + B) = ½⇒ cos (A + B) = cos 60º
⇒ A + B = 60º——-(ii)
Solving (i) and (ii) A = 45º; B = 15º
1 (b) Find the value of tan 2A
tan 2A = tan 30º =1/√3

Question 5

The teacher asked the students to correctly complete the following sentence about the rhombus.
“A rhombus has a side length of l units and one of its angles is equal to Ɵ. The ratio of the length of the diagonals is dependent on _________.”
Ashima : only l.
Bilal :only Ɵ
Chris : both l and Ɵ
Duleep : neither l nor Ɵ.
Who answered the question correctly ? Show your work and give valid reason

Solution

ratio of the diagonals BD/AC is only dependent on Ɵ not l.
Bilal answered correctly.

Question 6

If cosecx – sinx = a³, secx – cosx = b³, then prove that a² b² (a² + b² ) = 1

Solution

L.H.S= a² b²(a²+ b²)

= 1= R.H.S

Question 7

Find the value of 4/3 tan²30° + sin²60° – 3cos²60° +3/4 tan²60° – 2tan²45°

Solution

= 25/36

Question 8

Find the value of 4(sin⁴30°+ cos⁴60°) – 3(cos²45° + sin²90°).

Solution

= -4

Question 9

In an acute angled triangle ABC , if sin (A+B-C) =1/2 and cos (B+C-A) = 1/✓2 Find A, B, C

Solution

A+ B+ C = 180
A + B = 180 – C
B + C = 180 – A
sin (A+B-C) = 1/2
A+B-C =30
180 – C – C = 30
2C =150
C = 75
cos (B+C-A) = 1/√2
B+C-A =45
180 – A – A = 45
2A = 135 A =67.5
A = 67.5 , B = 37.5, C =75

Question 10

If A and (4A – 60°) are angles where A< 90°and (4A-60°) such that tan A = cot (4A-60°), then find sec A and cosec A?

Solution

sec 30°=2/√3 and Cosec30°= 2