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I want to find $angle BAD$. [closed]

-1

2

$begingroup$

$ angle BDC = 120°$, $angle BEC = 160°$. I want to find $angle BAD$.

geometry angle

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edited Apr 1 at 12:17

Anirban Niloy

8511319

asked Mar 30 at 9:24

BobtrollstenBobtrollsten

7115

$endgroup$

closed as off-topic by José Carlos Santos, Aretino, Xander Henderson, Shailesh, YiFan Apr 2 at 0:52

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – José Carlos Santos, Aretino, Xander Henderson, Shailesh, YiFan

If this question can be reworded to fit the rules in the help center, please edit the question.

add a comment | 

-1

2

$begingroup$

$ angle BDC = 120°$, $angle BEC = 160°$. I want to find $angle BAD$.

geometry angle

share|cite|improve this question

edited Apr 1 at 12:17

Anirban Niloy

8511319

asked Mar 30 at 9:24

BobtrollstenBobtrollsten

7115

$endgroup$

closed as off-topic by José Carlos Santos, Aretino, Xander Henderson, Shailesh, YiFan Apr 2 at 0:52

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – José Carlos Santos, Aretino, Xander Henderson, Shailesh, YiFan

If this question can be reworded to fit the rules in the help center, please edit the question.

add a comment | 

$begingroup$

$ angle BDC = 120°$, $angle BEC = 160°$. I want to find $angle BAD$.

geometry angle

share|cite|improve this question

edited Apr 1 at 12:17

Anirban Niloy

8511319

asked Mar 30 at 9:24

BobtrollstenBobtrollsten

7115

$endgroup$

share|cite|improve this question

edited Apr 1 at 12:17

Anirban Niloy

8511319

asked Mar 30 at 9:24

BobtrollstenBobtrollsten

7115

share|cite|improve this question

edited Apr 1 at 12:17

Anirban Niloy

8511319

asked Mar 30 at 9:24

BobtrollstenBobtrollsten

7115

edited Apr 1 at 12:17

Anirban Niloy

8511319

edited Apr 1 at 12:17

Anirban Niloy

8511319

asked Mar 30 at 9:24

BobtrollstenBobtrollsten

7115

asked Mar 30 at 9:24

BobtrollstenBobtrollsten

7115

1 Answer
1

active

oldest

votes

2

$begingroup$

Let $alpha = angle BAD = angle CAD$, $beta = angle ABD = angle EBD$, $gamma = angle ACD = angle ECD$.

The total of the angles in the quadrilateral $BACE,$ is
$$
2alpha + 2beta + 2gamma + 200^circ = 360^circ,
$$
so
$$
alpha + beta + gamma = 80^circ.
$$
The total of the angles in quadrilateral $BDCE,$ is
$$
beta + 120^circ + gamma + 200^circ = 360^circ,
$$
so
$$
beta + gamma = 40^circ.
$$
Subtracting this from the previous result, $alpha = 40^circ$.

share|cite|improve this answer

answered Mar 30 at 9:41

FredHFredH

3,7401024

$endgroup$

add a comment | 

2

$begingroup$

Let $alpha = angle BAD = angle CAD$, $beta = angle ABD = angle EBD$, $gamma = angle ACD = angle ECD$.

The total of the angles in the quadrilateral $BACE,$ is
$$
2alpha + 2beta + 2gamma + 200^circ = 360^circ,
$$
so
$$
alpha + beta + gamma = 80^circ.
$$
The total of the angles in quadrilateral $BDCE,$ is
$$
beta + 120^circ + gamma + 200^circ = 360^circ,
$$
so
$$
beta + gamma = 40^circ.
$$
Subtracting this from the previous result, $alpha = 40^circ$.

share|cite|improve this answer

answered Mar 30 at 9:41

FredHFredH

3,7401024

$endgroup$

add a comment | 

2

$begingroup$

Let $alpha = angle BAD = angle CAD$, $beta = angle ABD = angle EBD$, $gamma = angle ACD = angle ECD$.

The total of the angles in the quadrilateral $BACE,$ is
$$
2alpha + 2beta + 2gamma + 200^circ = 360^circ,
$$
so
$$
alpha + beta + gamma = 80^circ.
$$
The total of the angles in quadrilateral $BDCE,$ is
$$
beta + 120^circ + gamma + 200^circ = 360^circ,
$$
so
$$
beta + gamma = 40^circ.
$$
Subtracting this from the previous result, $alpha = 40^circ$.

share|cite|improve this answer

answered Mar 30 at 9:41

FredHFredH

3,7401024

$endgroup$

add a comment | 

$begingroup$

Let $alpha = angle BAD = angle CAD$, $beta = angle ABD = angle EBD$, $gamma = angle ACD = angle ECD$.

The total of the angles in the quadrilateral $BACE,$ is
$$
2alpha + 2beta + 2gamma + 200^circ = 360^circ,
$$
so
$$
alpha + beta + gamma = 80^circ.
$$
The total of the angles in quadrilateral $BDCE,$ is
$$
beta + 120^circ + gamma + 200^circ = 360^circ,
$$
so
$$
beta + gamma = 40^circ.
$$
Subtracting this from the previous result, $alpha = 40^circ$.

share|cite|improve this answer

answered Mar 30 at 9:41

FredHFredH

3,7401024

$endgroup$

share|cite|improve this answer

answered Mar 30 at 9:41

FredHFredH

3,7401024

answered Mar 30 at 9:41

FredHFredH

3,7401024

answered Mar 30 at 9:41

FredHFredH

3,7401024