Question

7. In Figure 5, ABCD is a rectangle. If ZCEB:ZECB = 3:2, find the measure of 1 angleECB 2 angleAEC
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Answer 1

Answer:

From the question it is given that,

From the question it is given that, ABCD is a rectangle∠CEB:∠ECB=3:2

From the question it is given that, ABCD is a rectangle∠CEB:∠ECB=3:2We have to find,

From the question it is given that, ABCD is a rectangle∠CEB:∠ECB=3:2We have to find,(i) ∠CEB and

From the question it is given that, ABCD is a rectangle∠CEB:∠ECB=3:2We have to find,(i) ∠CEB and(ii) ∠DCF

From the question it is given that, ABCD is a rectangle∠CEB:∠ECB=3:2We have to find,(i) ∠CEB and(ii) ∠DCFConsider the ΔBCE

From the question it is given that, ABCD is a rectangle∠CEB:∠ECB=3:2We have to find,(i) ∠CEB and(ii) ∠DCFConsider the ΔBCE∠B=90∘

From the question it is given that, ABCD is a rectangle∠CEB:∠ECB=3:2We have to find,(i) ∠CEB and(ii) ∠DCFConsider the ΔBCE∠B=90∘

From the question it is given that, ABCD is a rectangle∠CEB:∠ECB=3:2We have to find,(i) ∠CEB and(ii) ∠DCFConsider the ΔBCE∠B=90∘ Therefore, ∠CEB+∠ECB=90∘

From the question it is given that, ABCD is a rectangle∠CEB:∠ECB=3:2We have to find,(i) ∠CEB and(ii) ∠DCFConsider the ΔBCE∠B=90∘ Therefore, ∠CEB+∠ECB=90∘

Let us assume the angles be 3 y and 2y

Let us assume the angles be 3 y and 2y3y+2y=90∘

Let us assume the angles be 3 y and 2y3y+2y=90∘

Let us assume the angles be 3 y and 2y3y+2y=90∘ 5y=90∘

Let us assume the angles be 3 y and 2y3y+2y=90∘ 5y=90∘

Let us assume the angles be 3 y and 2y3y+2y=90∘ 5y=90∘ y=90∘/5

Let us assume the angles be 3 y and 2y3y+2y=90∘ 5y=90∘ y=90∘/5y=18°

Let us assume the angles be 3 y and 2y3y+2y=90∘ 5y=90∘ y=90∘/5y=18°

Let us assume the angles be 3 y and 2y3y+2y=90∘ 5y=90∘ y=90∘/5y=18° Then, ∠CEB=3y=3×18=54∘

Let us assume the angles be 3 y and 2y3y+2y=90∘ 5y=90∘ y=90∘/5y=18° Then, ∠CEB=3y=3×18=54∘

Let us assume the angles be 3 y and 2y3y+2y=90∘ 5y=90∘ y=90∘/5y=18° Then, ∠CEB=3y=3×18=54∘ ∠CEB=∠ECD

Let us assume the angles be 3 y and 2y3y+2y=90∘ 5y=90∘ y=90∘/5y=18° Then, ∠CEB=3y=3×18=54∘ ∠CEB=∠ECD54° =54° ... [alternate angles are equal]

We know that, sum of linear pair angles equal to 180°

We know that, sum of linear pair angles equal to 180°

We know that, sum of linear pair angles equal to 180° ∠ECD+DCF=180 ∘

We know that, sum of linear pair angles equal to 180° ∠ECD+DCF=180 ∘

We know that, sum of linear pair angles equal to 180° ∠ECD+DCF=180 ∘ 54∘+∠DCF=180°

We know that, sum of linear pair angles equal to 180° ∠ECD+DCF=180 ∘ 54∘+∠DCF=180°

We know that, sum of linear pair angles equal to 180° ∠ECD+DCF=180 ∘ 54∘+∠DCF=180° By transposing we get, ∠DCF=180° −54°

We know that, sum of linear pair angles equal to 180° ∠ECD+DCF=180 ∘ 54∘+∠DCF=180° By transposing we get, ∠DCF=180° −54°

We know that, sum of linear pair angles equal to 180° ∠ECD+DCF=180 ∘ 54∘+∠DCF=180° By transposing we get, ∠DCF=180° −54° ∠DCF=126°