Construct a triangle ABC in which angle B= 60^o,angle C=75^o  
and the perpendicular from the vertex A to the base BC is 4.5 cm
 

Dear Student,

Please find below the solution to the asked query:


 Given : , $\angle$ B = 60$°$ and $\angle$ C = 75$°$

Using angle sum property of triangle, we have

$\angle$ B + $\angle$ C + $\angle$ A  = 180$°$

60$°$ + 75$°$ + $\angle$ A  = 180$°$

$\angle$ A  = 45$°$

Steps of Construction:

Step 1 :  Draw a line BX .

Step 2 : Place the compass on point B and draw an angle of 60$°$ as shown.

Step 3 : Now, open the arc of compass equivalent to 4.5 cm and from point B and from any other point on BX draw an arc at P and Q as shown.

Step 4 : Join P and Q. Now, PQ is a line parallel to BX at an height of 4.5 cm from it.

Step 5 : The point where line BY and PQ meet each other is the point A.

Step 6 : From point A, take the angular bisector of 60$°$ and 120$°$ and draw an angle of 90$°$ as shown by dotted line AZ ( As $\angle$ BAZ  = 90$°$ .

Step 7 : Again,take the angular bisector of 90$°$ and draw an angle of 45$°$ as shown $\angle$ BAC = $\angle$ CAZ = 45$°$.Now, the point at which it cuts the BX is point C.

Hence, triangle ABC is the required triangle.

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