Geometry: Mensuration, Trigonometry

Question: tan³60° - 2sin60° = ?

Solution:
Given that,
tan³60° - 2sin60°
= (√3)³ - 2(√3/2)
= 3√3 - √3
= 2√3

Question: If θ be a positive acute angle satisfying cos²θ + cos⁴θ = 1, then the value of tan²θ + tan⁴θ is?

Solution: 
Given that, 
cos²θ + cos⁴θ = 1
⇒ cos⁴θ = 1 - cos²θ
⇒ cos⁴θ = sin²θ ; [sin²θ = 1 - cos²θ] 
⇒ cos²θ.cos²θ = sin²θ
⇒ cos²θ = sin²θ/cos²θ
⇒ cos²θ = tan²θ

Now, 
cos²θ + cos⁴θ = 1
⇒ cos²θ + (cos²θ)² = 1
⇒ tan²θ + (tan²θ)² = 1
∴ tan²θ + tan⁴θ = 1

Question: The length of a rectangular plot is 10 meters less than three times its breadth. If the cost of fencing the plot at Tk 50 per meter is Tk 15000, what is the length of the plot in meters?

Solution:
Let the breadth of the plot be x meters.
Then, the length of the plot is 3x - 10 meters.

Perimeter of the rectangle = 2 × (Length + breadth)
= 2 × (x + 3x - 10 )
= 2 × (4x - 10 )
= 8x - 20

Given,
Cost of fencing per meter = Tk 50
Total cost = Tk 15000

So, Perimeter × 50 = 15000
⇒ (8x - 20) × 50 = 15000
⇒ 8x - 20 = 15000/50
⇒ 8x = 300 + 20
⇒ 8x = 320 
∴ x = 40

∴ Breadth = 40 meters.
and length = 3x - 10 = 3 × 40 - 10 = 110 meters.

l = (48 - 16)m = 32 m, [because 8+8 = 16]
b = (36 -16)m = 20 m,
h = 8 m.
Volume of the box = (32 x 20 x 8) m cube
= 5120 m cube.

Question: If θ be an acute angle and 7sin²θ + 3cos²θ = 4, then the value of tanθ is?

Solution:
7sin²θ + 3cos²θ = 4
⇒ 7sin²θ + 3(1 - sin²θ) = 4
⇒ 7sin²θ + 3 - 3sin²θ = 4
⇒ 4sin²θ = 1
⇒ sin²θ = 1/4
⇒ sinθ = 1/2
⇒ sinθ = sin30°
∴ θ = 30°

∴ tanθ = tan30° = 1/√3

Given, Length of rectangle, L = 13 cm
breadth, B = 104/13 = 8 cm
So, its perimeter = 2(13+8) = 42
ATQ, 2πr = 130 - 42
Or, 2πr = 88
Or, r = 14
∴ Area of the circle = πr² = 22/7 × 14² = 616 cm²

Question: A square and a circle have the same perimeter. The side of the length of square is 44 cm, what is the area of the circle?

Solution:
Perimeter of the square = 4 × side length
= 4 × 44 cm
= 176 cm

As per the question, the square and circle have the same perimeter.
∴ Circumference of the circle = 176 cm
We know that, Circumference of the circle = 2πr
∴ 2πr = 176
⇒ r = 176 / (2π)
⇒ r = 88 / π
⇒ r = 88 / (22/7)
⇒ r = 88 × 7 / 22
⇒ r = 4 × 7
⇒ r = 28 cm

Area of the circle = πr²
= (22/7) × 28²
= (22/7) × (28 × 28)
= 22 × 4 × 28
= 2464 sq. cm

∴ The area of the circle is 2464 sq. cm.

Question: The midpoint of the line joining (10, 2) and (4, 8) is -

Solution:
The formula for the midpoint of (x₁, y₁) and (x₂, y₂) is
= ((x₁ + x₂)/2 , (y₁ + y₂)/2)

∴ The midpoint of the line joining (10, 2) and (4, 8) is
= (14/2 , 10/2)
= (7, 5)

Question: A rectangular field will be fenced on three sides, leaving one side of 20 feet uncovered. If the area of the field is 600 square feet, how many feet of fencing is required?

Solution:
আয়তাকার মাঠের ক্ষেত্রফল = 600 বর্গ ফুট
যে পাশে বেড়া দেওয়া হবে না তার দৈর্ঘ্য = 20 ফুট
অতএব, আয়তাকার মাঠের অন্য পাশের দৈর্ঘ্য = ক্ষেত্রফল/একপাশের দৈর্ঘ্য
= 600 / 20 = 30 ফুট

চতুর্দিকে বেড়া দেওয়ার প্রয়োজন নেই, কারণ একপাশ উন্মুক্ত থাকবে।
যে তিনটি পাশে বেড়া দিতে হবে, তাদের দৈর্ঘ্য হবে (30 + 20 + 30) ফুট।
সুতরাং, প্রয়োজনীয় বেড়ার মোট দৈর্ঘ্য = 30 + 20 + 30 = 80 ফুট।

Question: In a trapezoid, the lengths of the two parallel bases are 12 and 20 units, respectively. If the height of the trapezoid is 5, find the area of the trapezoid.

Solution:
Given that,
Trapezoid with bases a = 12 and b = 20
Height, h = 5

We know,
Area of trapezoid = (1/2) × (sum of bases) × height
= (1/2) × (a + b) × h
= (1/2) × (12 + 20) × 5
= (1/2) × 32 × 5
= 80

So, the area of the trapezoid is 80 square units.

Question: If the volume of a sphere is 2304π cm³, what is the surface area of the sphere?

Solution:
দেওয়া আছে,
গোলকের আয়তন, V = 2304π cm³

আমরা জানি,
গোলকের আয়তন, V = (4/3)πr³
⇒ (4/3)πr³ = 2304π
⇒ r³ = 2304 × (3/4)
⇒ r³ = 576 × 3
⇒ r³ = 1728
⇒ r = 12 সেমি

এখন,
গোলকের সমগ্র পৃষ্ঠতলের ক্ষেত্রফল, A = 4πr²
⇒ A = 4π(12)²
⇒ A = 4π × 144
⇒ A = 576π cm²

অতএব, গোলকের সমগ্র পৃষ্ঠতলের ক্ষেত্রফল হলো 576π cm²।

Question: A boy of height 1.5 m is walking away from the base of a lamp post at a speed of 0.8 m/sec. Find the height of the lamp post from the ground, if the shadow of the boy is 2.0 m after walking for 4 sec.

Solution:

Given that,
Height of the boy = 1.5 m
Speed of the boy = 0.8 m/s
Distance travelled by boy in 4 sec = 0.8 × 4 = 3.2 m
Total distance of shadow of boy and distance from base of lamp post = 2.0 + 3.2 = 5.2 m

Let the height of lamp post be 'h' m
According to question,
⇒ 1.5/2.0 = h/5.2
⇒ h = (5.2 × 1.5)/2.0
⇒ h = 3.9 m

So, The height of the lamp post is 3.9 meters.

Question: If tanθ =√3 and sinθ = √3/2. Find the value of cosθ.

Solution:
Given,
tanθ =√3 and sinθ = √3/2

We know,
tanθ = sinθ/cosθ
⇒√3 = (√3/2)/cosθ
⇒ cosθ = (√3/2)/√3
∴ cosθ = 1/2   

Let length = x and breadth = y. Then,
2(x + y)= 46 or
x + y = 23 and
x² + y² = (17)²
= 289.

Now, (x + y)² = (23)²
⇒ (x² + y²) + 2xy = 529
⇒ 289 + 2xy = 529
⇒ xy = 120.

∴ Area = xy = 120 cm²

Question: ∠P and ∠Q are complementary to each other. If ∠P = 20° + 4x and ∠Q = 6x, find the value of ∠Q.

Solution:
Here,
∠P = 20° + 4x and ∠Q = 6x

For complementary angles,
∠P + ∠Q = 90°
⇒ (20° + 4x) + 6x = 90°
⇒ 20° + 4x + 6x = 90°
⇒ 20° + 10x = 90°
⇒ 10x = 90° - 20°
⇒ 10x = 70°
∴ x = 7°

So, ∠Q = 6 × 7° = 42°

None of the above
Radius of the circle = r
So hypotenuse = 2r
So, 2r = √(2.4² + 1.8²)
Or, 2r = √9
Or, 2r = 3
the circumference of the circle = 2πr = 3π

Question: A 50-meter cable is attached from the top of a vertical pole down to the ground. If the cable makes an angle of 30 degrees with the ground, find the height of the pole.

Solution: 

ধরি,
উচ্চতা(Height), AB = h

দেয়া আছে,
AC = 50m                    
∠ACB = 30°

∴ sin30°= AB/AC
⇒ 1/2 = h/50
⇒ h = 50 × 1/2
∴ h = 25 m

Question: A ladder is leaning against a wall. It makes a 60° angle with the wall. If the distance between foot of ladder and wall is 7.5 meters, find the length of the ladder.

Solution:

Let BC be the wall and AC be the ladder.
∠BAC = 60° and AB = 7.5 meter
In ΔABC,
cos60° = AB/AC
⇒ 1/2 = 7.5/AC
⇒ AC = 7.5 × 2
∴ AC = 15

Total volume of water displaced = (4 x 50) m³ = 200 m³.
Rise in water level =200/(40 x 20)m 0.25 m = 25 cm.

Question: If tanθ = 5/12, then find the value of sinθ.

Solution:
Given, tanθ = Perpendicular/Base = 5/12
Here, base = 12 and perpendicular = 5
Let hypotenuse = x

Now, according to Pythagorean theorem,
x² = 5² + 12²
⇒ x² = 25 + 144
⇒ x = √169
⇒ x = 13

∴ sinθ = Perpendicular/Hypotenuse = 5/13

Question: The slope of the line 4x - 2y = 8 is not the same as the slope of which one of the following lines?
​
​Solution:
​প্রথমে, প্রদত্ত রেখাটির ঢাল নির্ণয় করতে হবে। 
​রেখাটির সমীকরণকে y = mx + c তে রূপান্তর করতে হবে। এখানে 'm' হলো ঢাল (Slope)।

প্রদত্ত রেখার সমীকরণ: 4x - 2y = 8
⇒ - 2y = - 4x + 8
⇒ y = (- 4/- 2)x + (8/- 2)
⇒ y = 2x - 4
∴ এই রেখাটির ঢাল (m) হলো 2.

এবার, প্রদত্ত অপশনগুলোর প্রত্যেকটির ঢাল নির্ণয় করি:
ক) 4x - 2y = 12
⇒ -2y = -4x + 12
⇒ y = 2x - 6
∴ ঢাল 2

খ) 2x - y = -5
⇒ -y = -2x - 5
⇒ y = 2x + 5
∴ ঢাল 2

গ) y = 2x - 1
∴ ঢাল 2

ঘ) x + 2y = 6
⇒ 2y = -x + 6
⇒ y = (-1/2)x + 3
∴ ঢাল - 1/2

সুতরাং, দেখা যাচ্ছে যে শুধুমাত্র অপশন (ঘ) এর রেখার ঢাল মূল রেখার ঢাল থেকে ভিন্ন।

Question: If y = sin(sinx) then what is the value of dy/dx?

Solution:

Question: If a right-angled isosceles triangle has base 6 cm, then height is:
(Officer Cash 2022 অনুযায়ী)

Solution:
(Right-angled isosceles triangle) সমকোণী সমদ্বিবাহু ত্রিভুজ এর ভূমি = 6 cm.
সমকোণী সমদ্বিবাহু ত্রিভুজ এর ভূমি ও উচ্চতা সমান।
ভূমি = উচ্চতা = 6 cm.
∴ উচ্চতা = 6 cm

প্রশ্ন: If θ be an acute angle and 5sin²θ + 3cos²θ = 4, then the value of tanθ is?

সমাধান:
5sin²θ + 3cos²θ = 4
⇒ 5sin²θ + 3(1 - sin²θ) = 4
⇒ 5sin²θ + 3 - 3sin²θ = 4
⇒ 2sin²θ = 1
⇒ sin²θ = 1/2
⇒ sinθ = √(1/2)
⇒ sinθ = 1/√2
⇒ sinθ = sin45°
⇒ θ = 45°

∴ tanθ = tan45° = 1

Area of the plot = (110 × 65)m²
= 7150m²
Area of the plot excluding the path = [(110 - 5) × (65 - 5)]m²
= 6300 m²

∴ Area of the path = (7150 - 6300) m²
= 850 m²

Cost of graveling the path = Tk. {850 × (80/100)}
= Tk. 680

Let original edge = a,
Then, original volume = a³

New edge = (150/100)a
= 3a/2
New volume = (3a/2)³
= 27a³/8

Increase in volume = (27a³)/8 - (a³)
= 19a³/8

∴ Increase% = {(19a³/8) × (1/a³) × 100}%
= 237.5%

Question: From the top of a lighthouse 60 m high above sea level, the angle of depression of a boat is 45°. How far is the boat from the foot of the lighthouse?

Solution:

Let the height of the lighthouse above sea be AC and it is given 60 m.
Angle of depression = 45°

Boat is at point B so the distance between the base of lighthouse A and Boat is AB.

 So, tan 45° = AC / AB
⇒ 1 = 60 / AB
⇒ AB = 60 m

∴ The boat is 60 m away from the foot of the lighthouse.

Question: How far apart are the centers of two circles with diameters of 16 cm and radii of 6 cm, when they touch each other externally? 

Solution:
আমরা জানি,
দুইটি বৃত্ত পরস্পরকে বহিঃস্পর্শ করলে কেন্দ্রদ্বয়ের মধ্যবর্তী দূরত্ব বৃত্ত দুইটির ব্যাসার্ধের যোগফলের সমান।

এখানে,
১ম বৃত্তের ব্যাসার্ধ = 16/2 = 8 সে.মি.
২য় বৃত্তের ব্যাসার্ধ = ৬ সে.মি.

∴ কেন্দ্রদ্বয়ের মধ্যবর্তী দূরত্ব = (8 + 6) সে.মি.
= 14 সে.মি.

Question: If θ lies in the first quadrant and cos²θ - sin²θ = 1/2. then the value of tan²2θ + sin²3θ is-

Solution:
cos²θ - sin²θ = 1/2
⇒ cos2θ = 1/2
⇒ cos2θ = cos60°
⇒ 2θ = 60°
⇒ θ = 30°

Now, tan²2θ + sin²3θ
= tan²60° + sin²90°
= 3 + 1
= 4

Question: If sec θ = 5/4, then what is the value of sinθ?

Solution:
এখানে,
secθ = 5/4 = অতিভুজ/ভূমি
∴ অতিভুজ = 5, ভূমি = 4

পিথাগোরাসের উপপাদ্য অনুসারে, লম্ব নির্ণয় করি,
লম্ব = √(অতিভুজ² - ভূমি²)
= √(5² - 4²)
= √(25 - 16)
= √9
= 3

এখন,
sinθ = লম্ব/অতিভুজ 
= 3/5

সুতরাং, sinθ = 3/5।

Question: If the side of a square is increased by 20%, by what percent will the area be increased?

Solution:
Let the original side length = 10 units.

∴ Area = 10 × 10 = 100 square units

Again, 
After a 20% increase, the new side length = 10 + (20% of 10)
= 10 + 2 = 12 units

∴ New area = 12 × 12 = 144 square units

∴ Increase in area = (144 - 100) square units
= 44 square units

∴ Percentage increase in area = (44/100) × 100%
= 44%

So the area will increase by 44%

Question: find 

Solution: 

Question: If C is the midpoint of the points A(2, - 1) and B(8, 5), find the length of AC.

Solution:
দেওয়া আছে,
 A(2, - 1) এবং B(8, 5), 
এবং C হলো AB-এর মধ্যবিন্দু।

দূরত্বের সূত্র ব্যবহার করে AB-এর দৈর্ঘ্য নির্ণয় করি,
AB = √(x₂ - x₁)² + (y₂ - y₁)²)
= √(8 - 2)² + {5 - (-1)}²
= √(6² + 6²)
= √(36 + 36)
= √72
= √(36 × 2)
= 6√2

C হলো AB এর মধ্যবিন্দু, তাই AC = AB/2
 = 6√2/2
 = 3√2

We know, Perimeter = 2(l+b) = 76
⇒ (l+b) = 38
⇒ l = 38 - b .....(1)
Given, the area, lb = 360 .....(2)
From (2)
(38 – b)b = 360
⇒ 38b – b² = 360
⇒ b² – 38b + 360 = 0
⇒ (b – 20)(b – 18) = 0
So, b = 20 or b = 18
When b = 20, l = 18 and when b = 18, l = 20.
∴ the shorter side is 18.