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slavikrds [6]
3 years ago
5

HELP PLEASE AND THANK YOU 4JAAB

Mathematics
1 answer:
Maslowich3 years ago
7 0

Answer:

\bold{Q13}\ \angle W\\\\\bold{Q14}\ \angle S\\\\\bold{Q15}\ side\ WA\\\\\bold{Q16}\ \triangle PTA

Step-by-step explanation:

For Q13, Q14, Q15

\text{for}\ \overline{WA}\ \text{and}\ \overline{AS}\ -\ A\ \text{is common}.\ \text{Therefore inclu}\text{ded angle is}\ \angle A\\\\\text{for}\ \overline{WS}\ \text{and}\ \overline{WA}\to\angle W\\\\\text{for}\ \overline{WS}\ \text{and}\ \overline{SA}\to\angle S\\\\============================

\text{Inc}\text{lude side between}\ \angle A\ \text{and}\ \angle W\ \text{is}\ AW\ (WA).\\\\\text{Inc}\text{lude side between}\ \angle S\ \text{and}\ \angle W\ \text{is}\ SW\ (WS).\\\\\text{Inc}\text{lude side between}\ \angle A\ \text{and}\ \angle S\ \text{is}\ AS\ (SA).

Q16

O\ \leftrightarrow\ T\\B\ \leftrightarrow\ A\\R\ \leftrightarrow\ P\\\\\text{Therefore}\ \triangle ROB\ \cong\ \triangle PTA

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