Question

Please solve the above problem


Thanks in advance !!!​

Answer 1

Answer:

2.9

Step-by-step explanation:

first measure this with help of compass system to inform you that you have received your answer is that the best .

Answer 2

$\large\boxed{\angle BAD=15^{\circ}}$

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In $\triangle ACD,$

$AC=CD\quad\implies\quad\angle CDA=\angle CAD\quad\longrightarrow\quad (1)$

But from $\triangle ABD,$ we get that,

$\angle BAD+\angle ABD=\angle CDA$

Because we know that the sum of two interior angles of a triangle is equal to the third exterior angle.

And,

$\angle CAD=\angle CAB-\angle BAD$

Then (1) becomes,

$\angle BAD+\angle ABD=\angle CAB-\angle BAD\\\\2\angle BAD=\angle CAB-\angle ABC\quad\quad [\angle ABD=\angle ABC]\\\\2\angle BAD=30^{\circ}\\\\\large\boxed{\angle BAD=15^{\circ}}$

Thus 15° is the answer.

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