Exercise 5.3
1. Find the value of the unknown 'x' in the following triangles.
(i)
img(src="/content/scert/7/triangles_properties/images/chapter4/q1i.png" width=220 height=200)
Solution:
In triangle ABC:
Sum of angles in a triangle =
Given angles =
Therefore, x = 180° - 50° - 60°
x =
(ii)
img(src="/content/scert/7/triangles_properties/images/chapter4/q1ii.png" width=220 height=200)
Solution:
In triangle PQR:
Given angle =
One angle is
Therefore, x = 180° - 90° - 30°
x =
(iii)
img(src="/content/scert/7/triangles_properties/images/chapter4/q1iii.png" width=220 height=200)
Solution:
In triangle XYZ:
Given angles =
Therefore, x = 180° - 20° - 110°
x =
2. Find the values of the unknowns 'x' and 'y' in the following diagrams.
(i)
img(src="/content/scert/7/triangles_properties/images/chapter4/q2i.png" width=220 height=200)
Solution:
In triangle PQR:
Given angles =
x + y + 50° + 120° =
x + y =
Also, x = y (by symmetry)
Therefore, x = y =
(ii)
img(src="/content/scert/7/triangles_properties/images/chapter4/q2ii.png" width=220 height=200)
Solution:
In triangle RST:
Given angles =
x + y + 50° =
x + y =
Also, x = y (by symmetry)
Therefore, x = y =
(iii)
img(src="/content/scert/7/triangles_properties/images/chapter4/q2iii.png" width=220 height=200)
Solution:
In triangle MNA:
Given angles =
x + y + 50° + 60° =
x + y =
Also, x = y (by symmetry)
Therefore, x = y =
(iv)
img(src="/content/scert/7/triangles_properties/images/chapter4/q2iv.png" width=220 height=200)
Solution:
In triangle ABC:
Given angles =
x + y + 30° + 60° =
x + y =
Also, x = y (by symmetry)
Therefore, x = y =
(v)
img(src="/content/scert/7/triangles_properties/images/chapter4/q2v.png" width=220 height=200)
Solution:
In triangle EFG:
Given angle =
x + y + 90° =
x + y =
Also, x = y (by symmetry)
Therefore, x = y =
(vi)
img(src="/content/scert/7/triangles_properties/images/chapter4/q2vi.png" width=220 height=200)
Solution:
Since the lines are parallel:
x + y =
Also, x = y (by symmetry)
Therefore, x = y =
3. Find the measure of the third angle of triangles whose two angles are given below:
(i)
38°, 102°
Solution:
Sum of angles in a triangle =
Given angles =
Third angle = 180° - 38° - 102°
Third angle =
(ii)
116°, 30°
Solution:
Sum of angles in a triangle =
Given angles =
Third angle = 180° - 116° - 30°
Third angle =
(iii)
40°, 80°
Solution:
Sum of angles in a triangle =
Given angles =
Third angle = 180° - 40° - 80°
Third angle =
4. In a right-angled triangle, one acute angle is 30°. Find the other acute angle.
Solution:
In a right-angled triangle:
One angle is
Given acute angle =
Sum of angles in a triangle =
Therefore, other acute angle = 180° - 90° - 30°
Other acute angle =
5. State true or false for each of the following statements.
(i)
A triangle can have two right angles.
Solution:
Sum of angles in a triangle =
Two right angles would be =
This leaves
Is this possible?
(ii)
A triangle can have two acute angles.
Solution:
An acute angle is less than
Example: A triangle with angles
All these angles are acute (less than 90°)
Is this possible?
(iii)
A triangle can have two obtuse angles.
Solution:
An obtuse angle is greater than
If we have two obtuse angles: Each angle > 90° Two angles total >
This exceeds sum of angles in a triangle
Is this possible?
(iv)
Each angle of a triangle can be less than 60°.
Solution:
If all angles are less than 60°: Maximum possible sum =
But sum must equal exactly 180°
Is this possible?
6. The angles of a triangle are in the ratio 1
3. Find the angles.
Solution:
Let's solve step by step:
Sum of angles in a triangle =
Let's say the parts are x, 2x, and 3x
Then: x + 2x + 3x =
6x = 180°
x =
Therefore: First angle =
7. In the figure, DE || BC, ∠A = 30° and ∠B = 50°. Find the values of x, y and z.
img(src="/content/scert/7/triangles_properties/images/chapter4/q7.png" width=220 height=200)
Solution:
Given: DE is parallel to BC
When lines are parallel: Corresponding angles are equal Alternate angles are equal
Given angles: ∠A =
Therefore: {.reveal(when="blank-2")}x =
8. In the figure, ∠ABD = 3∠DAB and ∠CDB = 96°. Find ∠ABD.
img(src="/content/scert/7/triangles_properties/images/chapter4/q8.png" width=220 height=200)
Solution:
Given: ∠CDB =
Let ∠DAB = x
Then ∠ABD =
In triangle ABD: Sum of angles =
x + 3x + 96° = 180°
4x + 96° = 180°
4x =
x =
Therefore: ∠ABD = 3 × 21° =
9. In ΔPQR, ∠P = 2∠Q and 2∠R = 3∠Q, calculate the angles of ΔPQR.
Solution:
Let's say ∠Q = x
Then:
∠P =
∠R =
In a triangle: Sum of angles =
2x + x + 1.5x = 180°
4.5x = 180°
x =
Therefore:
∠Q = 40°
∠P = 2 × 40° =
∠R = 1.5 × 40° =
10. If the angles of a triangle are in the ratio 1
5, find the angles.
Solution:
Let's solve step by step:
Sum of angles in a triangle =
Let's say the smallest part is x
Then:
First angle = x
Second angle =
Third angle =
Using sum of angles:
x + 4x + 5x = 180°
10x = 180°
x =
Therefore:
First angle = 18°
Second angle = 4 × 18° =
Third angle = 5 × 18° =
11. The acute angles of a right triangle are in the ratio 2:3. Find the angles of the triangle.
Solution:
In a right triangle:
One angle is
Sum of all angles =
Sum of acute angles = 180° - 90° =
Let's say the parts are 2x and 3x
Then:
2x + 3x = 90°
5x = 90°
x =
Therefore:
First acute angle = 2 × 18° =
Second acute angle = 3 × 18° =
The three angles are: 90°, 36°, and 54°
12. In the figure, ΔPQR is right angled at Q, ML || RQ and ∠LMR = 130°. Find ∠MPL, ∠LMP and ∠QRP.
img(src="/content/scert/7/triangles_properties/images/chapter4/q12.png" width=220 height=200)
Solution:
Given:
∠LMR =
ML is parallel to RQ
∠Q =
Since ML || RQ:
∠MPL =
∠LMP = 180° - 130° =
∠QRP =
13. In Figure ABCDE, find ∠1 + ∠2 + ∠3 + ∠4 + ∠5.
img(src="/content/scert/7/triangles_properties/images/chapter4/q13.png" width=220 height=200)
Solution:
Let's solve step by step:
At any point:
Sum of angles around a point =
In the figure:
∠1 + ∠2 + ∠3 + ∠4 + ∠5 make a complete rotation
Therefore:
∠1 + ∠2 + ∠3 + ∠4 + ∠5 =