Question

In the figure DE is parallel to QR , AP and BP are bisector of <EAB and <RBA respectively. find <APB.

Answer 1

Here is the answer of your que.

Answer 2

Answer:

HEY MATE YOUR ANSWER IS HERE

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IN TRIANGLE PQR

AC||PR (GIVEN)

HENCE

\frac{qc}{cp} = \frac{qa}{ar}

cp

qc

=

ar

qa

(by BPT THEOREAM)

hence

\frac{15}{cp } = \frac{12}{20}

cp

15

=

20

12

NOW ...

CP =

\frac{15}{12} \times 20

12

15

×20

hence

CP = 25cm -----------eq 1

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NOW IN TRIANGLE PQR AGAIN.....

CB || QR ( GIVEN )

HENCE

\frac{pc}{qc} = \frac{pb}{br}

qc

pc

=

br

pb

( BY BPT THEOREAM )

HENCE

from eq 1 value of PC = 25 cm

\frac{25}{15} = \frac{15}{br}

15

25

=

br

15

HENCE BR =

\frac{15}{25} \times 15

25

15

×15

hence

BR = 9 cm

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THANKS FOR THE QUESTION

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