Properties of Triangles — Chapter Test | Class 11 IMO

Q1

The angle of elevation of the top of a tower from a point A due south of the tower is 30°. From a point B due east of A, the angle is 15°. If AB = 40m, finding the tower's height requires setting up right triangles and applying:

Let height be h. Distance from south point A is h cot30° = h√3. The point B is due east of A, so the triangle formed by the base of the tower, A, and B on the ground is a right-angled triangle at A. The distance from B to base is h cot15°. The Pythagoras theorem is applied to this ground triangle to link h and AB=40m.

Q2

The length of the shadow of a vertical tower increases by 10 meters when the altitude of the sun changes from 45° to 30°. What is the height of the tower?

Let height be h. Initial shadow = x. tan45° = h/x → x = h. New shadow = x + 10. tan30° = h/(x+10) → 1/√3 = h/(h+10). So, h√3 = h + 10, h(√3 − 1) = 10, h = 10 / (√3 − 1). Rationalizing, h = 10(√3 + 1) / 2 = 5(√3 + 1) m.

Q3

If the sides of a triangle are 3, 5, and 7, what is the measure of its greatest angle?

The greatest angle is opposite the longest side (7). Let this angle be C. cosC = (3² + 5² − 7²)/(2 × 3 × 5) = (9 + 25 − 49)/30 = −15/30 = −1/2. Therefore, C = 120°.

Q4

In a ∆ABC, a cosA + b cosB + c cosC is equal to:

Using Sine Rule, a = 2R sinA, etc. Expression becomes 2R(sinA cosA + sinB cosB + sinC cosC) = R(sin2A + sin2B + sin2C). For angles of a triangle, sin2A + sin2B + sin2C = 4 sinA sinB sinC. So the result is 4R sinA sinB sinC.

Q5

A surveyor measures a triangular plot. Two sides are 50m and 80m. The angle between them is measured as 60°. Using the Cosine Rule, what is the length of the third side?

Let sides be a=50, b=80, and angle C=60°. c² = 50² + 80² − 2(50)(80)cos60° = 2500 + 6400 − 8000(1/2) = 8900 − 4000 = 4900. c = √4900 = 70 m.

Q6

In a triangle, if the sides are in Arithmetic Progression (A.P.), then the product of the inradius and the exradius opposite to the middle side is equal to:

Let sides be a, b, c in AP. Then a+c = 2b. s = (a+b+c)/2 = 3b/2. r = ∆/s and exradius r₂ = ∆/(s-b). Product r × r₂ = ∆² / (s(s-b)). Since s = 3b/2, s-b = b/2. Thus s(s-b) = 3b²/4. This simplifies to ∆² / (s(s-b)).

Q7

In ∆ABC, if a = 18, b = 24, c = 30, find the radius of the incircle (r).

Sides 18, 24, 30 are a multiple of the 3-4-5 right triangle. Area ∆ = ½ × 18 × 24 = 216. Semi-perimeter s = (18 + 24 + 30)/2 = 36. Inradius r = ∆ / s = 216 / 36 = 6.

Q8

In ∆ABC, if a = 3, b = 4, and c = 5, what is the value of sin(A + B)?

Since the sides are 3, 4, 5, angle C is 90°. In any triangle, A + B + C = 180°, so A + B = 180° − 90° = 90°. Therefore, sin(A + B) = sin(90°) = 1.

Q9

A triangle has sides 10, 10 and 10. Its area is:

Area of equilateral triangle = √3/4 × 10² = 25√3.

Q10

If a = b = c, then each angle measures:

An equilateral triangle has all angles 60°.

Q11

If a, b, c are the sides of a triangle and a² + b² < c², then the angle C is:

By Cosine Rule, cosC = (a² + b² − c²)/(2ab). If a² + b² < c², the numerator is negative, making cosC negative. This means angle C lies between 90° and 180°, so it is obtuse.

Q12

The side opposite the smallest angle in a triangle is:

Angles and opposite sides increase together.

Q13

If a circle passes through all three vertices of a right-angled triangle with sides 6, 8, and 10, what is the area of this circle?

The circle passing through all three vertices is the circumcircle. For a right-angled triangle, the circumradius R is half the hypotenuse. Hypotenuse = 10, so R = 5. The area of the circle is πR² = π(5²) = 25π.

Q14

In ∆ABC, the value of (b² − c²) / a² is equal to:

By Sine Rule, replace sides with proportional sines: (sin²B − sin²C) / sin²A. We know sin²B − sin²C = sin(B+C)sin(B−C). Also, sin(B+C) = sin(180°−A) = sinA. So, (sinA sin(B−C)) / sin²A = sin(B−C) / sinA.

Q15

In ∆ABC, 2(bc cosA + ca cosB + ab cosC) is equal to:

From the Cosine Rule, 2bc cosA = b² + c² − a², 2ca cosB = c² + a² − b², and 2ab cosC = a² + b² − c². Adding these three equations together results in (b² + c² − a²) + (c² + a² − b²) + (a² + b² − c²) = a² + b² + c².