Properties of Triangles MCQ Quiz in मल्याळम - Objective Question with Answer for Properties of Triangles - സൗജന്യ PDF ഡൗൺലോഡ് ചെയ്യുക

Formula used :

sin 2θ = 2 sin θ cos θ 

cos C + cos D = 2cos(\(\frac{C+D}{2}\)) cos(\(\frac{C-D}{2}\))

Calculaton :

sin A - cos B - cos C = 0

⇒ sin A = cos B + cos C

Using the formula given above

⇒ 2 sin(A/2) cos(A/2) = 2cos(\(\frac{B+C}{2}\)) cos(\(\frac{B-C}{2}\))

We know that, for a ΔABC, ∠ A + ∠ B + ∠ C = π 

⇒ sin(A/2) cos(A/2) = cos(\(\frac{π - A}{2}\)) cos(\(\frac{B-C}{2}\))

Since, cos(π/2 - θ) = sin θ 

⇒ sin(A/2) cos(A/2) =  sin(\(\frac{ A}{2}\)) cos(\(\frac{B-C}{2}\))

⇒ cos(A/2) =  cos(\(\frac{B-C}{2}\))

⇒ A/2 = (B - C)/2

⇒ A = B - C

⇒ B = A + C

But ∠ A + ∠ B + ∠ C = π 

⇒ B = π - B

⇒ 2B = π 

⇒  B = π/2

∴ Angle  B is equal to \(\frac{\pi}{2}\).

Concept:

• The triangle inequality states that for any triangle, the sum of the lengths of any two sides must be greater than or equal to the length of the remaining side.

Calculation:

Given: a, b, c are the sides of a triangle ABC

Since, a, b, c are positive numbers.

⇒ a^1/p, b^1/p, c^1/p are positive numbers.     ...(∵ p > 1)

We know that sum of any two sides must be greater than third side.

⇒ a^1/p + b^1/p > c^1/p

⇒ a^1/p + b^1/p - c^1/p > 0

Concept:

Angle sum property: The sum of the angles of a triangle is 180°.

Types of triangles:

• Acute angle triangle: If all the angles of a triangle are less than 90° then it is an acute angle triangle
• Obtuse angle triangle: If any one of the angles of a triangle is greater than 90° then it is an obtuse angle triangle.
• Right angle triangle: If one angle of a triangle is 90° and the side opposite to 90° is the longest side then it is a right angle triangle.
• Equilateral Triangle: If all the angles of a triangle are 60° then it is an equilateral triangle.

Calculation:

Given: In Δ ABC ∠A = y, ∠B = x and ∠C = y + 20 and 2x – y = 20.

As we know that sum of all the angles of triangle = 180°

y + x + y + 20 = 180

x + 2y = 160      ----(1)

2x – y = 20      ----(2)

From (1) and (2)

x = 40° and y = 60°

The angles are 40°, 60° and 80°

As we can see that all angles of Δ ABC are less than 90° 

Concept:

Cosine Rule:

a² = b² + c² - 2bc cosA

b² = a² + c² - 2ac cosB

c² = b² + a² - 2ab cosC

The formula can be rearranged to:

\(cosA=\frac{b^{2}+c^{2}-a^{2}}{2bc}\)

\(cosB=\frac{a^{2}+c^{2}-b^{2}}{2ac}\)

\(cosC=\frac{a^{2}+b^{2}-c^{2}}{2ab}\)

Solution:

Given, In a triangle ABC

b = 2, c = 3 and ∠B = \(\frac{\pi }{6}\)

By using Cosine Rule in ΔABC,

\(cosB=\frac{a^{2}+c^{2}-b^{2}}{2ac}\)

⇒ \(cos\frac{\pi }{6}=\frac{a^{2}+3^{2}-2^{2}}{2.a.3} \)

⇒ \(\frac{\sqrt{3}}{2}=\frac{a^{2}+5}{6a} \)

⇒ \(a^{2}-3\sqrt{3}a+5=0\)

∴ The correct option is (3)

∠C = 54° and ∠EAC = 42°, So ∠AEC = 84° (sum of all angles of triangle = 180°)

∴ ∠BEA = 180 - 84 = 96°.

∵ DE bisects AB perpendicularly, so angle made by BD and DA should be equal.

∴ ∠DBE + ∠BED + ∠EDB = 180°

∴ ∠DBE + 96/2 + 90 = 180

∴ ∠ABC = ∠DBE = 180 - 90 - 48 = 42°

CONCEPT:

Properties of Triangles:

If three sides a, b and c are sides of a triangle, \({\rm{\Delta }} = {\rm{\;}}\sqrt {{\rm{s}}\left( {{\rm{s}} - {\rm{a}}} \right)\left( {{\rm{s}} - {\rm{b}}} \right)\left( {{\rm{s}} - {\rm{c}}} \right)} \)where \({\rm{s}} = \frac{{{\rm{a}} + {\rm{b}} + {\rm{c}}}}{2}\) where

S = semi perimeter of a triangle

Half angle formulae of a triangle:

 In Δ ABC

• \(\sin \left( {\frac{{\rm{A}}}{2}} \right) = {\rm{\;}}\sqrt {\frac{{({\rm{s}} - {\rm{b}}(\left( {{\rm{s}} - {\rm{c}}} \right)}}{{{\rm{bc}}}}} \)
• \(\cos \left( {\frac{{\rm{A}}}{2}} \right) = {\rm{\;}}\sqrt {\frac{{{\rm{s}}\left( {{\rm{s}} - {\rm{a}}} \right)}}{{{\rm{bc}}}}} \)
• \(\tan \left( {\frac{{\rm{A}}}{2}} \right) = {\rm{\;}}\sqrt {\frac{{\left( {{\rm{s}} - {\rm{b}}} \right)\left( {{\rm{s}} - {\rm{c}}} \right)}}{{{\rm{s}}\left( {{\rm{s}} - {\rm{a}}} \right)}}} \)

CALCULATION:

Given that, a – 2b + c = 0

⇒ a + c = 2b

As we know that, S (Semi perimeter) of a triangle = (a + b + c)/2

⇒ s = (2b + b) /2 = 3b/2

As we know that, \({\rm{Cot\;}}\left( {\frac{{\rm{A}}}{2}} \right) = \sqrt {\frac{{{\rm{s}}\left( {{\rm{s}} - {\rm{a\;}}} \right)}}{{\left( {{\rm{s}} - {\rm{b}}} \right)\left( {{\rm{s}} - {\rm{c}}} \right)}}} ,\cot \left( {\frac{{\rm{c}}}{2}} \right) = {\rm{\;}}\sqrt {\frac{{{\rm{s}}\left( {{\rm{s}} - {\rm{c}}} \right)}}{{\left( {{\rm{s}} - {\rm{a}}} \right)\left( {{\rm{s}} - {\rm{b}}} \right)}}} \)

\(\Rightarrow \cot \left( {\frac{{\rm{A}}}{2}} \right)\cot \left( {\frac{{\rm{c}}}{2}} \right) = {\rm{\;}}\sqrt {\frac{{{\rm{s}}\left( {{\rm{s}} - {\rm{a\;}}} \right)}}{{\left( {{\rm{s}} - {\rm{b}}} \right)\left( {{\rm{s}} - {\rm{c}}} \right)}}} \sqrt {\frac{{{\rm{s}}\left( {{\rm{s}} - {\rm{c}}} \right)}}{{\left( {{\rm{s}} - {\rm{a}}} \right)\left( {{\rm{s}} - {\rm{b}}} \right)}}} \)

\(\Rightarrow \cot \left( {\frac{{\rm{A}}}{2}} \right)\cot \left( {\frac{{\rm{c}}}{2}} \right) = {\rm{\;}}\sqrt {\frac{{{{\rm{s}}^2}}}{{{{\left( {{\rm{s}} - {\rm{b}}} \right)}^2}}}} \)

\(\Rightarrow \cot \left( {\frac{{\rm{A}}}{2}} \right)\cot \left( {\frac{{\rm{c}}}{2}} \right) = \frac{{\rm{s}}}{{\left( {{\rm{s}} - {\rm{b}}} \right)}}\)

Concept:

If A, B and C are angles and a, b and c are the sides of a ΔABC, then: \(\frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C~}\)

sin (x + y) = sin x ⋅ cos y + sin y ⋅ sin x

Calculation:

Let us consider Δ ABC with A, B and C as its angles and a, b and c as its sides.

Given: Angles are in AP i.e A, B and C are in AP.

⇒ 2 × B = A + C

As we know that, A + B + C =180°

⇒ 3 × B = 180°     ----(∵ 2 × B = A + C)

⇒ B = 60°

As we know that, if A, B and C are angles and a, b and c are the sides of a Δ ABC, then:

\(\frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C~}\)

\(⇒ \frac{b}{c}=\frac{\sin B}{\sin C}\)

As it is given that b : c = √3 : √2 and sin C = sin 60° = √ 3 / 2.

\(⇒ \frac{√{3}}{√{2}}=\frac{\frac{√{3}}{2}}{\sin C}~⇒ \sin C=\frac{1}{√{2}}\)

⇒ C = 45°

As we know that, sin (x + y) = sin x ⋅ cos y + sin y ⋅ sin x

⇒ sin (B + C) = sin B ⋅ cos C + sin C ⋅ cos B

⇒ sin (B + C) = √3 / 2 ⋅ 1 / √2 + 1 / √2 ⋅ 1 / 2 = 1 + √3 / 2√2

Given:  

In ΔABC, ∠B = 90°, AC = 169 cm, and BC = 120 cm

Concept used: 

ABC is a right angled triangle at B

Pythagoras theorem = AC² = AB² + BC²

Calculation:

BC = 120 cm, AC = 169 cm

According to Pythagoras theorem, 

⇒ (169)²  = AB² + (120)²

⇒ AB²  = 14161

⇒ AB = 119 cm

∴ Option 3 is correct.

Concept:

Pythagoras theorem: In a right-angled triangle, the square of the hypotenuse side is equal to the sum of squares of the other two sides.

Sine Rule in triangle ABC, having sides a, b, c:

\(\rm \frac{\sin A}{a}=\frac{\sin B}{b}=\frac{\sin C}{c}=\frac{1}{K}\)

sin A - cos B = \(\rm 2\cos(\frac{A +B}{2})\sin(\frac{A -B}{2})\)

Calculation:

1. We have,

sin \(\rm B = \frac 1 3=\frac{P}{H}\)

Hwence, Right angle triangle can be drawn as,

Here, P = 1
H = 3
B = x

Using Pythagoras theorem,

3² = 1² + x²

x² = 8

\(\therefore x=2\sqrt{2}\)

Now , Cosec C = \(\frac{3}{2\sqrt{2}}\)

Hence Ststement (1) is not correct.

2. Given,

b cos B = c cos C

Using the sine rule,

\(\rm \frac{\sin A}{a}=\frac{\sin B}{b}=\frac{\sin C}{c}=\frac{1}{K}\)

⇒ b = K sin B & c = K sin C

⇒ 2 sin B cos B = 2 sin C cos C

Assume,
K = 2

⇒ 2 sin B cos B = 2 sin C cos C

⇒ sin 2B = sin 2C  [2 sin x cos x = sin 2x]

⇒ sin 2B - sin 2C = 0

Using Formula:

sin C - sin D = 2 cos \((\frac{C+D}{2})\) × sin \((\frac{C-D}{2})\)

⇒ 2 cos (B + C) sin (B - C) = 0

In this case,

Either,

cos (B + C) = 0

Hence, (B + C) = 90° .... (1)

Or,

Or sin (B -C) = 0

B - C = 0

⇒ B = C  .... (2)

From equation (1) & (2),

B = C = 45°

So ABC must be issosceles 

Hence, option (2) is correct.

Concept:

The sum of the angles of the triangle is 180º.

Cosine formula: cos(B) = \(\rm \dfrac{a^2+c^2-b^2}{2ac}\)

Calculations:

Given, in a triangle ABC if the angles A, B, C are in AP.

⇒2B = A+ C

We know that, the sum of the angles of the triangle is 180º.

⇒ A + B + C = 180º

 ⇒ 2B + B = 180º

⇒ 3B = 180º

⇒B =  60º

By cosine formula, we have

cos(B) = \(\rm \dfrac{a^2+c^2-b^2}{2ac}\)

⇒cos(60º) = \(\rm \dfrac{a^2+c^2-b^2}{2ac}\)

⇒\(\rm\dfrac{1}{2}\) = \(\rm \dfrac{a^2+c^2-b^2}{2ac}\)

⇒\(\rm a^2+c^2-b^2= ac\)

⇒b2 = a2 + c2 - ac

Hence, In a triangle ABC if the angles A, B, C are in AP, b² = a² + c² - ac is the correct statement.