Question

tell me 4 question. ​

Answer 1

Solution :-

1)

Given that tan θ = 8/15

On squaring both sides then

=> (tan θ)² = (8/15)²

=> tan² θ = 64/225

On adding 1 both sides then

=> tan² θ + 1 = (64/225)+1

=> tan² θ + 1 = (64+225)/225

=> tan² θ + 1 = 289/225

We know that

sec² θ - tan² θ = 1

=> sec² θ = 289/225

=> sec θ = √(289/225)

=> sec θ = 17/15

Therefore, sec θ = 17/15

2)

In ∆ PQR , ∠ PQR = 90°

QS ⊥ PR

QS = 12 cm

QS = 12 cmRS = 16 cm

We know that

∆PQS and ∆QSR and ∆ PQR are similar triangles

So , we know that

The square of the perpendicular drawn from the right angle to the hypotenuse is equal to the product of the two sides on the hypotenuse formed by the perpendicular.

Therefore, QS² = PS×SR

=> 12² = PS × 16

=> 144 = PS × 16

=> PS = 144/16

=> PS = 9 cm

Now,

We have ,

PR = PS + SR

=> PR = 9+16

Therefore, PR = 25 cm

The length of the side PR = 25 cm

3)

Given that

D lies on x-axis

We know that

The equation of x-axis is y = 0

So ,The ordinate of D = 0

The distance between D from y-axis

= 5 units

The distance from y-axis = abscissa = 5

The coordinates of D = (5,0)

4)

Given that

The ratio of two triangles with common base = 4:5

Let they be 4X cm² and 5X cm²

The area of the bigger triangle = 5X cm²

The area of the smaller triangle = 4X cm²

Given that

The base of two triangles are same

Let the base of the two triangles be b cm

Height of the smaller triangle = 6 cm

Now,

Area of the triangle = (1/2)×base×height sq.units

=> Area of the smaller triangle

=> (1/2)×b×6 = 4

=> 6b/2 = 4

=> 3×b = 4

=> b= 4/3 cm

Therefore, b = 4/3 cm

The length of the base = 4/3 cm

Area of the greater triangle = 5 cm²

=> (1/2)×(4/3)×h = 5

=> (4/6)×h = 5

=> (2/3)×h = 5

=> h = 5/(2/3)

=> h = 5×3/2

=> h = 15/2

=> h = 7.5 cm

Therefore, Height of the greater triangle is 7.5 cm

Answer 2

Step-by-step explanation:

Solution :-

1)

Given that tan θ = 8/15

On squaring both sides then

=> (tan θ)² = (8/15)²

=> tan² θ = 64/225

On adding 1 both sides then

=> tan² θ + 1 = (64/225)+1

=> tan² θ + 1 = (64+225)/225

=> tan² θ + 1 = 289/225

We know that

sec² θ - tan² θ = 1

=> sec² θ = 289/225

=> sec θ = √(289/225)

=> sec θ = 17/15

Therefore, sec θ = 17/15

2)

In ∆ PQR , ∠ PQR = 90°

QS ⊥ PR

QS = 12 cm

QS = 12 cmRS = 16 cm

We know that

∆PQS and ∆QSR and ∆ PQR are similar triangles

So , we know that

The square of the perpendicular drawn from the right angle to the hypotenuse is equal to the product of the two sides on the hypotenuse formed by the perpendicular.

Therefore, QS² = PS×SR

=> 12² = PS × 16

=> 144 = PS × 16

=> PS = 144/16

=> PS = 9 cm

Now,

We have ,

PR = PS + SR

=> PR = 9+16

Therefore, PR = 25 cm

The length of the side PR = 25 cm

3)

Given that

D lies on x-axis

We know that

The equation of x-axis is y = 0

So ,The ordinate of D = 0

The distance between D from y-axis

= 5 units

The distance from y-axis = abscissa = 5

The coordinates of D = (5,0)

4)

Given that

The ratio of two triangles with common base = 4:5

Let they be 4X cm² and 5X cm²

The area of the bigger triangle = 5X cm²

The area of the smaller triangle = 4X cm²

Given that

The base of two triangles are same

Let the base of the two triangles be b cm

Height of the smaller triangle = 6 cm

Now,

Area of the triangle = (1/2)×base×height sq.units

=> Area of the smaller triangle

=> (1/2)×b×6 = 4

=> 6b/2 = 4

=> 3×b = 4

=> b= 4/3 cm

Therefore, b = 4/3 cm

The length of the base = 4/3 cm

Area of the greater triangle = 5 cm²

=> (1/2)×(4/3)×h = 5

=> (4/6)×h = 5

=> (2/3)×h = 5

=> h = 5/(2/3)

=> h = 5×3/2

=> h = 15/2

=> h = 7.5 cm

Therefore, Height of the greater triangle is 7.5 cm.

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