Question

1) If sec (x²-6x +80)° = cosec (40-5x)°, find the value of x.
2) If sin (β+70°) = cos(50°-α),then find the value of sin(α-β).
3) Find the simplified value of sin43°sec47°-cot41°cot49°.
4) The value of sin42°cos48°+cos42°sin48° is one of the following:
a)0 b)1 c)sin6° d)cos6°
5) In ∆ABC, ∠ABC= 90°, AB=6cm,BC = 8cm.If a perpendicular BD is drawn from B on AC, find the length of BD. ​

Answer 1

Answer:

Question :-

1) If sec (x² - 6x + 80)° = cosec (40 - 5x)°, find the value of x.

Solution :-

Since, sec (x² - 6x + 80)° = cosec (40 - 5x)°

➪ (x² - 6x + 80)° + (40 - 5x)° = 90°

➪ x² - 6x + 80 + 40 - 5x - 90 = 0

➪ x² - 6x - 5x + 80 + 40 - 90 = 0

➪ x² - 11x + 120 - 90 = 0

➪ x² - 11x + 30 = 0

➪ x² - (6 + 5)x + 30 = 0

➪ x² - 6x - 5x + 30 = 0

➪ x(x - 6) - 5(x - 6) = 0

➪ (x - 6) (x - 5) = 0

➪ (x - 6) = 0

➪ x - 6 = 0

➪ x = 6

Either,

➪ (x - 5) = 0

➪ x - 5 = 0

➪ x = 5

∴ The value of x is 6, 5

Question :-

2) If sin (β + 70°) = cos (50° - α), then find the value of sin (α - β).

Solution :-

Since, sin (β + 70°) = cos (50° - α)

➪ sin (β + 70°) = sin [90° - (50° - α)]

➪ β + 70° = 90° - (50° - α)

➪ β + 70° - 90° + 50° - α = 0

➪ β - α + 30° = 0

➪ α - β = 30°

∴ sin (α - β)

➪ sin 30°

➪ 1/2

∴ The value of sin (α - β) is 1/2

Question :-

3) Find the simplified value of sin 43° sec 47° - cot 41° cot 49°.

Solution :-

➪ sin 43° sec 47° - cot 41° cot 49°

➭ sin 43° sec (90° - 43°) - cot 41° cot (90° - 41°)

➭ sin 43° cosec 43° - cot 41° tan 41°

➭ sin 43° × 1/sin 43° - cot 41° × 1/cot 41°

➭ 1 - 1

➭ 0

∴ The value of sin 43° sec 47° - cot 41° cot 49° is 0

Question :-

4) The value of sin 42° cos 48° + cos 42° sin 48° is one of the following:

a) 0 b) 1 c) sin 6° d) cos 6°.

Solution :-

➪ sin 42° cos 48° + cos 42° sin 48°

➭ sin 42° cos (90° - 42°) + cos 42° sin (90° - 42°)

➭ sin 42° sin 42° + cos 42° cos 42°

➭ sin² 42° + cos² 42°

➭ 1

∴ (b) is correct

Question :-

5) In ∆ ABC, ∠ABC = 90°, AB = 6 cm, BC = 8 cm. If a perpendicular BD is drawn from B on AC, find the length of BD.

Given :-

• In ∆ ABC , ∠ABC = 90°, AB = 6cm, BC = 8 cm. If a perpendicular BD is drawn from B on AC,

Find Out :-

• The length of BD

Solution :-

In ∆ ABC, by Pythagoras theorem,

AC²

➪ AB² + BC²

➪ 6² + 8²

➪ 6 × 6 + 8 × 8

➪ 36 + 64

➪ 100

➪ 10²

∴ AC

➪ 10²

Now, if AD = x

DC

➪ AC - AD

➪ 10 - x

and, BD = y

Now, in ∆ ABC,

➪AD² + BD²

➭ AB²

➪ x² + y² = 6²

➭ x² + y² = 6 × 6

∴ x² + y² = 36. . . .(i)

Again, in ∆ BDC,

➪ BD² + DC²

➭ BC²

➪ y² + (10 - x)² = 8²

➭ y² + (10 - x)² = 8 × 8

∴ y² + (10 - x)² = 64

➪ y² + {(10)² - 2 × 10 × x + (x)²} = 64

∴ y² + 100 - 20x + x² = 64. . . .(ii)

Subtracting (i) from (ii)

➪ (y² + 100 - 20x + x²) - (x² + y²)

➭ 64 - 36

➭ 28

or,

➪ 20x = 100 - 28

➪ 20x = 72

➪ x = 72/20

➪ x = 3.6

Putting in (i), (3.6)² + y² = 36

➪ y² = 36 - (3.6)²

➪ y² = 36 - 12.96

➪ y² = 23.04

➪ y² = (4.8)

➪ y = 4.8

∴ The length of BD = 4.8 cm

Formula used :-

➪ (a - b)²

➭ a² - 2ab + b²