RBSE Solutions Class 9 Maths Chapter 8 Construction of Triangles Exercise 8.3

Chapter 8 Construction of Triangles Ex 8.3

Case III: To construct a triangle when a side and two angles are given.

Question 1. Construct a triangle PQR when QR = 8 cm, \( \angle Q = 120^\circ \) and \( \angle R = 30^\circ \).
Answer: We need to draw a triangle PQR following the given measurements. First, draw the base QR. Then, at each end of the base, draw the given angles, making sure they meet to form the third vertex, P. The sum of angles in a triangle is always \( 180^\circ \).
Steps of construction:
1. Draw a line segment QR with a length of 8 cm.
2. From point Q, draw a ray making an angle of \( 120^\circ \) with QR.
3. From point R, draw another ray making an angle of \( 30^\circ \) with QR. This ray should intersect the first ray at a point, which we will call P.
Hence, \( \triangle PQR \) is the triangle we needed to construct.

In simple words: Draw a straight line for the base. At one end, make a wide angle (120 degrees). At the other end, make a smaller angle (30 degrees). Where these two angle lines cross each other, that's the top point of your triangle.

🎯 Exam Tip: Always make a rough sketch first to visualize the triangle and plan your construction steps, especially for angles larger than 90 degrees.

Question 2. Construct a triangle ABC when b = 7 cm, \( \angle A = 90^\circ \) and \( \angle C = 60^\circ \).
Answer: For this construction, we have a side (b) and two angles (\( \angle A \) and \( \angle C \)). First, we need to find the third angle, \( \angle B \). We know that \( \angle A + \angle B + \angle C = 180^\circ \). So, \( 90^\circ + \angle B + 60^\circ = 180^\circ \), which means \( \angle B = 180^\circ - 150^\circ = 30^\circ \). Side b is opposite angle B, so b = AC = 7 cm. We will draw side AC first. Then we will draw angles A and C at its ends.
Steps of construction:
1. Draw a line segment AC with a length of 7 cm.
2. At point A, draw a ray making an angle of \( 90^\circ \) with AC.
3. At point C, draw another ray making an angle of \( 60^\circ \) with AC. This ray should intersect the first ray at a point, which we will call B.
4. Join B to C.
Hence, \( \triangle ABC \) is the right-angled triangle we needed to construct. This triangle has a right angle at A, making it a right-angled triangle.

In simple words: First, calculate the missing angle. Then, draw the side whose length is given. At each end of that side, draw the known angles. Where the lines from these angles meet, that's the third point of your triangle.

🎯 Exam Tip: When given two angles and a side (ASA criterion), always calculate the third angle first. This helps in understanding the triangle's overall shape and properties.

Question 3. Construct an isosceles triangle whose base is of length 4 cm and the vertex angle is equal to \( 30^\circ \). Draw a perpendicular from the vertex to the base.
Answer: In an isosceles triangle, the two base angles are equal. We know the sum of angles in a triangle is \( 180^\circ \). So, if the vertex angle is \( 30^\circ \), the sum of the two base angles is \( 180^\circ - 30^\circ = 150^\circ \). Therefore, each base angle is \( \frac{150^\circ}{2} = 75^\circ \).
So, we have a base BC = 4 cm, and \( \angle B = \angle C = 75^\circ \).
Steps of construction:
1. Draw a line segment BC with a length of 4 cm to serve as the base.
2. At point B, draw a ray making an angle of \( 75^\circ \) with BC.
3. At point C, draw another ray also making an angle of \( 75^\circ \) with BC. This ray should intersect the first ray at a point, which we will call A.
4. From vertex A, draw a perpendicular line down to the base BC. This line will divide the isosceles triangle into two right-angled triangles.
Hence, \( \triangle ABC \) is the required isosceles triangle with base 4 cm and vertex angle \( 30^\circ \).

In simple words: For a triangle with two equal sides, if you know the top angle, subtract it from 180 degrees, then divide the answer by two to find the two bottom angles. Draw the base, then make the two bottom angles from each end. Where they meet is the top point. Draw a line straight down from the top point to the base.

🎯 Exam Tip: Remember that in an isosceles triangle, the angles opposite the equal sides are also equal. This property is key for finding unknown angles and constructing the triangle.