Construct $△$ CXV such that CX = 9.1 cm , $∠$ CVX = 130 $°$ ,VD $⊥$ CX and VD = 1.7 cm.

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Solution

Steps of construction:
(i) Draw a line segment CX = 9.1 cm.
(ii) Through C, draw CY such that $∠$ XCY= 112 $°$ .
(iii) Draw CZ $⊥$ CY.
(iv) Draw the perpendicular bisector of CX intersecting CZ at O and CX at H.
(v) With O as the centre and OC as the radius, draw a circle.
(vi) The minor arc CWX of the circle contains the vertical 130 $°$ .
(vii) With H as the centre, cut off HP = 2.4 cm.
(viii) Draw a line VV' parallel to CX passing through P.
(ix) With V as the centre, draw a line parallel to PH intersecting XY at D.
(x) Join VC and VX.

Thus, $△$ CVX or $△$ CV'X is the required triangle.

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Construct △CXV such that CX = 9.1 cm , ∠CVX = 130° ,VD⊥CX and VD = 1.7 cm.

Construct $△$ CXV such that CX = 9.1 cm , $∠$ CVX = 130 $°$ ,VD $⊥$ CX and VD = 1.7 cm.