HOW TO CONSTRUCT A TRIANGLE PQR SUCH THAT PQ =5 cm ANGLE PQR =105 DEGREE &ANGLE QRP=30 DEGREE?

FAST TOMORROW IS MY EXAM!!!!!

Hi Kriti!
 
It is know that the sum of all interior angles of a triangle is 180°
In ∆PQR ;∠PQR + ∠QPR + ∠QRP = 180°
⇒ 105° + ∠QPR + 30° = 180°
⇒ 135° + ∠QPR = 180°
⇒ ∠QPR = 180° – 135° = 45°
 
In order to draw the ∆PQR draw PQ of length 5 cm. At P drawn an angle XPQ whose measure is 45°. At Q drawn an angle YQP whose measure is 105°, let the rays PX and QY interest R. Now, ∆PQR is the required triangle
 
 
Hope! This will help you.
Cheers!

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thanks

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