Very Short Answers on Properties of Triangles Class 11 OP Malhotra Exe-7B ISC Maths Solutions Ch-7. In this article you would learn to solve all questions on Properties of Triangles. Step by step solutions of latest textbook has been given as latest syllabus. Visit official Website CISCE for detail information about ISC Board Class-11

Properties of Triangles Class 11 OP Malhotra  Very Short Answer ISC Maths Solutions Ch-7

Board	ISC
Publications	S Chand
Subject	Maths
Class	11th
Chapter-7	Properties of Triangle.
Writer	OP Malhotra
Exe-7(B)	Very Short Answer Type Questions.

 Very Short Answers on Properties of Triangles

OP Malhotra ISC Class 11 Maths Solutions

Que-1: Find the area of a Δ ABC in which a = 1, b = 2 and ∠C = 60°.

Sol:  Area formula: (1/2)ab sinC
= (1/2)(1)(2) sin60°
= 1 × (√3/2)
= √3/2 sq.units

Que-2: In a Δ ABC, BC = √39, AC = 5 and AB = 7. What is the measure of ∠A?

Sol: Using Cosine Rule:
BC² = AB² + AC² − 2(AB)(AC)cosA
39 = 49 + 25 − 70cosA
39 = 74 − 70cosA
70cosA = 35
cosA = 1/2
∠A = 60°

Que-3: Two sides of a triangle are given by the roots of the equation x² − 5x + 6 = 0 and the angle between the sides is π/3. Find the perimeter of the triangle.

Sol:  Solve equation:
x² − 5x + 6 = 0
(x − 2)(x − 3) = 0
Sides = 2 and 3
Using Cosine Rule for third side:
c² = 2² + 3² − 2(2)(3)cos(π/3)
= 4 + 9 − 12(1/2)
= 13 − 6 = 7
c = √7
Perimeter = 2 + 3 + √7
= (5 + √7)

Que-4: In a Δ ABC, if ∠A = 120°, b = 2, ∠C = 30°, then a is equal to _____.

Sol:  B = 180° − (120° + 30°) = 30°
Using sine rule,
a/sin120° = 2/sin30°
a/(√3/2) = 2/(1/2) = 4
a = 4 × (√3/2) = 2√3

Que-5: In a Δ ABC, ∠A = 30°, b = 8, a = 6, then ∠B = sin⁻¹(x). Find x.

Sol:  Using sine rule,
a/sinA = b/sinB
6/sin30° = 8/sinB
6/(1/2) = 8/sinB
12 = 8/sinB
sinB = 8/12 = 2/3
x = 2/3

Que-6: In a Δ ABC, show that cosec A (sin B cos C + cos B sin C) = 1.

Sol:  sinB cosC + cosB sinC = sin(B + C)
= sin(180° − A) = sinA
Thus, cosecA × sinA = 1

Que-7: In a Δ ABC, show that a (b cos C − c cos B) = b² − c².

Sol:  Using cosine rule:
b cosC = (a² + b² − c²)/(2a)
c cosB = (a² + c² − b²)/(2a)
Subtracting,
b cosC − c cosB = (b² − c²)/a
Multiplying by a:
a(b cosC − c cosB) = b² − c²

Que-8: If in a Δ ABC, 4 sin A = 4 sin B = 3 sin C, then find the value of cos C.

Sol:  4sinA = 3sinC
⇒ sinA = (3/4)sinC
Similarly sinB = (3/4)sinC
So A = B
Thus C = 180° − 2A
Using sinA = (3/4)sinC and solving,
cosC = −1/8

Que-9: Find the area of the triangle whose sides are 6, 5, √13.

Sol:  Using Heron’s Formula:
s = (a + b + c)/2 = (6 + 5 + √13)/2
Area = √[s(s−6)(s−5)(s−√13)]
After simplification,
Area = 6 square units.

Que-10: If ∠A = 30°, c = 7√3 and ∠C = 90° in the Δ ABC, then find the value of a.

Sol: Since ∠C = 90°, side c is hypotenuse.
Using sin A = a / c
sin 30° = a / (7√3)
1/2 = a / (7√3)
a = (7√3)/2

–: End Properties of Triangles Class 11 OP Malhotra Exe-7B ISC Maths Ch-7 Solutions :–

Return to :- OP Malhotra ISC Class-11 S Chand Publication Maths Solutions
Thanks
Please share with your friends