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    				| 1. | In `DeltaP Q R ,`right-angled at Q (see figure), `P Q = 3 c m a n d P R = 6 c m`. Determine `/_Q P R`and `/_P R Q` | 
| Answer» If we create a right angle triangle `Delta PQR` with given details, then, `sinR = PQ/PR = 3/6 = 1/2 =sin30^@=> R =30^@` We know, `/_P+/_Q+/_R = 180^@` `/_P+90+30 = 180=>/_P = 180-120=60^@` So,`/_QPR = 60^@ and /_PRQ = 30^@` | |