Exercise 5.4
1. In ΔABC, name all the interior and exterior angles of the triangle.
img(src="/content/scert/7/triangles_properties/images/chapter4/q4-1.png" width=220 height=200)
Solution:
Let's identify the angles:
Interior angles of ΔABC are:
Exterior angles of ΔABC are:
At vertex A:
At vertex B:
At vertex C:
2. For ΔABC, find the measure of ∠ACD.
img(src="/content/scert/7/triangles_properties/images/chapter4/q4-2.png" width=220 height=200)
Solution:
Given in the figure:
∠CAB =
∠ABC =
In ΔABC:
Sum of angles =
∠BCA = 180° - (66° + 45°)
∠BCA =
∠ACD is an
∠ACD = 180° - ∠
∠ACD =
3. Find the measure of angles x and y.
img(src="/content/scert/7/triangles_properties/images/chapter4/q4-3.png" width=220 height=200)
Solution:
Given in the figure:
∠C =
∠CBD =
In ΔABC:
Sum of angles =
x + 30° + 65° = 180°
x = 180° - 95°
Therefore, x =
For angle y:
y + 65° = 180° (linear pair)
Therefore, y =
4. In the following figures, find the values of x and y.
img(src="/content/scert/7/triangles_properties/images/chapter4/q4-4a.png" width=220 height=200)
img(src="/content/scert/7/triangles_properties/images/chapter4/q4-4b.png" width=220 height=200)
Solution:
For the first figure:
Given: ∠A =
In ΔABC:
Sum of angles =
107° + x + 57° = 180°
x = 180° - 164°
Therefore, x =
For the second figure:
Given: ∠ADE =
y = ∠CDE - ∠ADE
y = 107° - 40°
Therefore, y =
5. In the figure ∠BAD = 3∠DBA, find ∠CDB, ∠DBC and ∠ABC.
img(src="/content/scert/7/triangles_properties/images/chapter4/q4-5.png" width=220 height=200)
Solution:
Given in the figure:
∠BAD =
∠DBC =
Given that ∠BAD = 3∠DBA:
Let ∠DBA = x
Then, 104° = 3x
Therefore, x =
∠CDB = 180° - 104° =
∠ABC = 26° + 65° =
6. Find the values of x and y in the following figures.
Figure (i)
img(src="/content/scert/7/triangles_properties/images/chapter4/q4-6a.png" width=320 height=200)
Solution:
Given: ∠A =
Due to equal marks, x = x
In triangle:
70° + x + x =
2x =
Therefore, x =
Figure (ii)
img(src="/content/scert/7/triangles_properties/images/chapter4/q4-6b.png" width=320 height=200)
Solution:
Given angle =
Due to equal marks, x = x
x + x + 50° =
2x =
Therefore, x =
Figure (iii)
img(src="/content/scert/7/triangles_properties/images/chapter4/q4-6c.png" width=320 height=200)
Solution:
Given angle =
Due to alternate angles:
x =
Due to corresponding angles:
y =
Figure (iv)
img(src="/content/scert/7/triangles_properties/images/chapter4/q4-6d.png" width=320 height=200)
Solution:
Given angle =
Due to equal marks, y = y
110° + y + y =
2y =
Therefore, y =
Figure (v)
img(src="/content/scert/7/triangles_properties/images/chapter4/q4-6e.png" width=320 height=200)
Solution:
Given: One angle =
Other angle =
Sum of angles =
Therefore:
x =
y =
Figure (vi)
img(src="/content/scert/7/triangles_properties/images/chapter4/q4-6f.png" width=320 height=200)
Solution:
Given angle =
Due to equal marks, x = x
80° + x + x =
2x =
Therefore, x =
7. One of the exterior angles of a triangle is 125° and the interior opposite angles are in the ratio 2:3. Find the angles of the triangle.
Solution:
Given: Exterior angle =
Let's find the interior angle:
Interior angle = 180° - 125° =
Let's say the opposite angles are 2x and 3x
Then:
2x + 3x = 55° (sum of opposite angles)
5x =
x =
Therefore:
First angle = 2 × 11° =
Second angle = 3 × 11° =
Third angle = 180° - (22° + 33°) =
8. The exterior ∠PRS of ΔPQR is 105°. If Q = 70°, find ∠P. Is ∠PRS > ∠P?
Solution:
Given:
Exterior ∠PRS =
∠Q =
Let's find interior ∠R:
Interior ∠R = 180° - 105° =
In ΔPQR:
∠P + 70° + 75° =
Therefore:
∠P = 180° - (70° + 75°) =
Comparing:
∠PRS = 105° and ∠P = 35°
Therefore,
9. If an exterior angle of a triangle is 130° and one of the interior opposite angle is 60°. Find the other interior opposite angle.
Solution:
Given:
Exterior angle =
One interior opposite angle =
Let's find the interior angle:
Interior angle = 180° - 130° =
The two interior opposite angles sum to the interior angle:
60° + x = 50°
Therefore:
Other interior opposite angle (x) =
Since angles cannot be negative:
This triangle is
10. One of the exterior angle of a triangle is 105° and the interior opposite angles are in the ratio 2:5. Find the angles of the triangle.
Solution:
Given:
Exterior angle =
Let's find the interior angle:
Interior angle = 180° - 105° =
Let's say the opposite angles are 2x and 5x
Then:
2x + 5x = 75° (sum of opposite angles)
7x =
x =
Therefore:
First angle = 2 × 15° =
Second angle = 5 × 15° =
Third angle = 180° - (30° + 75°) =