Question

In the given figure, AD ll BC. If angle ADB=5/4 angle BDC and angle BDC=4/7angle DCB, then find the value of angle BCD, angle ADB and angle DAB ​

Answer 1

Answer:

Step-by-step explanation:

Answer 2

Given:

AD // BC

∠ADB = 5/4 ∠BDC  

∠BDC = 4/7∠DCB  

To find:  

∠BCD, ∠ADB and ∠DAB

Solution:

We have,

$\angle ADB = \frac{5}{4} \angle BDC$  

So, we can assume

∠ADB = 5x° and ∠BDC = 4x°

Similarly, we have

$\angle BDC = \frac{4}{7} \angle DCB$

So, we can assume

∠BDC = 4x° and ∠DCB = 7x°

∵ AD // BC  

∴ ∠ADB = ∠DBC = 5x° ..... [alternate angles]

From the given figure, we can say,

In Δ BCD, using the angle sum property, we get

∠BDC + ∠DCB + ∠DBC = 180°

substituting the values

⇒ 4x + 7x + 5x = 180°

⇒ 16x = 180°

⇒ x = 11.25°

Therefore,

∠BCD = 7x° = 4 × 11.25° = 78.75°

and

∠ADB = 5x° = 5 × 11.25° = 56.25°

Now, in Δ ABD, we have

∠DAB + ∠ADB + ∠ABD = 180° ..... [angle sum propety]

substituting the values of ∠ADB = 70° (as given in the figure) & ∠ABD = 56.25°

⇒ ∠DAB + 70° + 56.25 = 180°  

⇒ ∠DAB = 180° - (70° + 56.25°)

⇒ ∠DAB = 53.75°

Thus, the value of the angles are as follows:

$\boxed{\bold{\angle BCD = \underline{78.75\°}}}\\\\\boxed{\bold{\angle ADB = \underline{56.25\°}}}\\\\\boxed{\bold{\angle DAB = \underline{53.75\°}}}$

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