Mohan has a triangle field ABC. He divided the whole field into two triangular fields ABD and ACD. After measuring he found that BC = AB, ∠EBC = 40⁰ and ∠CAD = 30⁰. Again, he divided the whole field into two triangular fields ABE and CBE. His son is in Class IX. So, he assumed ∠ACD = x⁰ and ∠ADB = y⁰. He prepared some questions based on his field and he asked his son to solve the questions: Answer the following questions: 41.
Find the value of x. (a) 50° (b) 60° (c) 70° (d) None of these 42.
Find the value of y. (a) 90° (b) 80° (c) 70° (d) None of these 43.
Find ∠ADC. (a) 90° (b) 110° (c) 120° (d) None of these 44.
Find ∠BAD. (a) 30° (b) 40° (c) 20° (d) None of these
Answers
Answer:
Case Study – 1
Mohan has a triangle field ABC. He divided the whole field into two triangular fields ABD and ACD.
After measuring he found that BC = AB, ∠EBC = 40⁰ and ∠CAD = 30⁰. Again, he divided the whole
field into two triangular fields ABE and CBE. His son is in Class IX. So, he assumed ∠ACD = x⁰ and
∠ADB = y⁰. He prepared some questions based on his field and he asked his son to solve the questions:
Answer the following questions:
41. Find the value of x.
(a) 50° (b) 60° (c) 70° (d) None of these
Ans. (a)
42. Find the value of y.
(a) 90° (b) 80° (c) 70° (d) None of these
Ans. (b)
43. Find ∠ADC.
(a) 90° (b) 110° (c) 120° (d) None of these
Ans. (d)
44. Find ∠BAD.
(a) 30° (b) 40° (c) 20° (d) None of these
Ans. (c)
45. Find ∠ABE.
(a) 30° (b) 40° (c) 20° (d) None of these
Ans. (b)
Given:
- Triangle ABC is first divided into two triangular fields Triangle ABD and triangle ACD.
- BC= AB
- ∠EBC = 40°
- ∠CAD=30°
- Triangle ABC is again divided to form two more triangular fields triangle ABE and triangle CBE
To find:
- ∠ACD = x°
- ∠ADB = y°
- ∠ADC
- ∠BAD
Solution:
We know that: Sum of all angles in a triangle = 180°
It is given BC= AB, so the triangle ABC is an isosceles triangle
∠ABC = ∠ACB (angles opposite to two equal sides of a triangle are also equal)
In triangle DAC, using exterior angle property we obtain:
∠DAC + ∠DCA= ∠ADB
30° + x° = y°
Thus the relation between x and y will be as follows:
y° - x° = 30°
From the given options we can see option (a) that is, 50° for the value of x, and option (b) that is, 80° for the value of y satisfy this relation.
Thus, x° = 50° (option a) and y°= 80° (option b)
Now, ∠ADC + x° = 180° (Straight line angle)
∠ADC = 180° - 50° = 130°
Since none of the options is = 130°, the answer for ∠ADC will be option(d) None of these.
Now, ∠x° = ∠BAC
∠BAC = 30° + ∠BAD
50° = 30° + ∠BAD
∠BAD= 50° - 30° = 20°
Thus, the ∠BAD= 20°, option (c) is the correct answer in this case.