Find the unknown angles x and y in the following given figures.

The given figure is, W Z ∥ Y X , W Y is a transversal. Therefore, ∠ Z W Y = W Y X and ∠ Y W X = ∠ W Y Z (Alternate interior angles are equal) ∴ ∠ Z W Y = 43 ° and ∠ Y W Z = 28 ° Now, in △ W O X ∠ W O X + ∠ O W X + ∠ O X W = 180 ° ⇒ x + 28 ° + 40 ° = 180 ° ⇒ x + 68 ° = 180 ° ⇒ x = 180 ° - 68 ° ⇒ x = 112 ° Also, V W X is a straight line. Therefore, ∠ Z W V + ∠ Z W X = 180 ° ⇒ y + 28 ° + 43 ° = 180 ° ⇒ y + 71 ° = 180 ° ⇒ y = 180 ° - 71 ° ⇒ y = 109 ° Hence, the value of x and y is 112 ° and 109 ° respectively.

The following figure shows a simple device in which A M B is a straight rod and C M is a rod pivoted at M with A M = B M = C M . Find ∠ A C B , if ∠ C A M = 60 ° , if ∠ A M C = 36 ° , and, if ∠ B M C = 100 ° . What can you generalise from the results from the above.

Given, ∠ C A M = 60 ° In △ A M C , A M = M C , So, ∠ A C M = ∠ C A M = 60 ° Also, ∠ A C M + ∠ C A M + ∠ A M C = 180 ° ⇒ ∠ A M C = 180 ° - 120 ° = 60 ° Now, A M B is a straight line. So, ∠ A M C + ∠ C M B = 180 ° ⇒ ∠ C M B = 180 ° - 60 ° = 120 ° Again, in △ C M B ∠ C M B + ∠ C B M + ∠ M C B = 180 ° ⇒ 120 ° + ∠ M C B + ∠ M C B = 180 ° ( C M = M B , s o , ∠ C B M = ∠ M C B ) ⇒ ∠ M C B = 60 ° 2 = 30 ° Thus, ∠ A C B = ∠ A C M + ∠ M C B ∠ A C B = 60 ° + 30 ° ∠ A C B = 90 ° Hence, ∠ A C B is equal to 90 ° . Given, ∠ A M C = 36 ° In △ A M C , A M = M C , So, ∠ A C M = ∠ C A M = x Also, ∠ A C M + ∠ C A M + ∠ A M C = 180 ° ⇒ 2 x = 180 ° - 36 ° = 144 ° ⇒ x = 144 ° 2 = 72 ° Now, A M B is a straight line. So, ∠ A M C + ∠ C M B = 180 ° ⇒ ∠ C M B = 180 ° - 36 ° = 144 ° Again, in △ C M B ∠ C M B + ∠ C B M + ∠ M C B = 180 ° ⇒ 144 ° + ∠ M C B + ∠ M C B = 180 ° ( C M = M B , s o , ∠ C B M = ∠ M C B ) ⇒ ∠ M C B = 36 ° 2 = 18 ° Thus, ∠ A C B = ∠ A C M + ∠ M C B ∠ A C B = 72 ° + 18 ° ∠ A C B = 90 ° Hence, ∠ A C B is equal to 90 ° . Given, ∠ B M C = 100 ° In △ C M B ∠ C M B + ∠ C B M + ∠ M C B = 180 ° ⇒ 100 ° + ∠ M C B + ∠ M C B = 180 ° ( C M = M B , s o , ∠ C B M = ∠ M C B ) ⇒ ∠ M C B = 80 ° 2 = 40 ° Now, A M B is a straight line. So, ∠ A M C + ∠ C M B = 180 ° ⇒ ∠ C M A = 180 ° - 100 ° = 80 ° Now, in △ A M C , A M = M C , So, ∠ A C M = ∠ C A M = x Also, ∠ A C M + ∠ C A M + ∠ A M C = 180 ° ⇒ 2 x = 180 ° - 80 ° = 100 ° ⇒ x = 100 ° 2 = 50 ° Thus, ∠ A C B = ∠ A C M + ∠ M C B ∠ A C B = 50 ° + 40 ° ∠ A C B = 90 ° Therefore, the value of ∠ A C B is 90 ° . Hence, we can generalise that ∠ A C B is always equal to 90 °

E M B I B E

CPA Maths - A Singapore Approach>Chapter 11 - Properties of Triangles>EXERCISE

Vistor Chow wal keung | RC Yadav Solutions for Chapter: Properties of Triangles, Exercise 1: EXERCISE

Author:Vistor Chow wal keung | RC Yadav

Vistor Chow wal keung | RC Yadav Mathematics Solutions for Exercise - Vistor Chow wal keung | RC Yadav Solutions for Chapter: Properties of Triangles, Exercise 1: EXERCISE

Attempt the free practice questions on Chapter 11: Properties of Triangles, Exercise 1: EXERCISE with hints and solutions to strengthen your understanding. CPA Maths - A Singapore Approach solutions are prepared by Experienced Embibe Experts.

Questions from Vistor Chow wal keung | RC Yadav Solutions for Chapter: Properties of Triangles, Exercise 1: EXERCISE with Hints & Solutions

MEDIUM

7th CBSE

IMPORTANT

CPA Maths - A Singapore Approach>Chapter 11 - Properties of Triangles>EXERCISE>Q 5

Find the unknown angles $x$ and $\Rightarrow x = 112 °$ in the following given figures.

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EASY

7th CBSE

IMPORTANT

CPA Maths - A Singapore Approach>Chapter 11 - Properties of Triangles>EXERCISE>Q 6

The figure below shows a section of a roof in which $A B = A C$ and $∠ A B C = 25 °$ . Find $∠ B A C$ .

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HARD

7th CBSE

IMPORTANT

CPA Maths - A Singapore Approach>Chapter 11 - Properties of Triangles>EXERCISE>Q 7

The figure below shows a rectangular field. $A C$ and $B D$ intersect at $E$ such that $A E = B E = C E = D E$ . If $∠ A E D = 64 °$ , find the angles $x, y$ and $z$ .