in a rhombus ABCD angle ACB is equal to 60 find (a) angle bac (b) angle dca (c) angle adc
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Given:- ∠ACB=60°
So, ∠ACB=∠BAC=60° [ angles opposite to equal side are equal ]
In triangle BAC:-
∠ACB+∠BAC+∠CBA=180° [ ANGLE SUM PROPERTY OF A TRAINGLE ]
60°+60°+∠CBA=180°
120°+∠CBA=180°
∠CBA=180°-120°
∠CBA=60°
We know that rhombus is a parallelogram, and opposite angles of a parallelogram are equal.
So,∠CBA=∠ADC=60° [ OPPOSITE ANGLES OF A PARALLELOGRAM ARE EQUAL AND RHOMBUS IS ALSO A PARALLELOGRAM ]
∠ACB=∠DAC=60° [ ALTERNATE INTERIOR ANGLES ARE EQUAL ]
In triangle ADC:-
∠ADC+∠DAC+∠DCA=180° [ ANGLE SUM PROPERTY OF A TRIANGLE ]
60°+60°+∠DCA=180°
120°+∠DCA=180°
∠DCA=180°-120°
∠DCA=60°
So, ∠DCA=60°
∠BAC=60°
∠ADC=60°
So, ∠ACB=∠BAC=60° [ angles opposite to equal side are equal ]
In triangle BAC:-
∠ACB+∠BAC+∠CBA=180° [ ANGLE SUM PROPERTY OF A TRAINGLE ]
60°+60°+∠CBA=180°
120°+∠CBA=180°
∠CBA=180°-120°
∠CBA=60°
We know that rhombus is a parallelogram, and opposite angles of a parallelogram are equal.
So,∠CBA=∠ADC=60° [ OPPOSITE ANGLES OF A PARALLELOGRAM ARE EQUAL AND RHOMBUS IS ALSO A PARALLELOGRAM ]
∠ACB=∠DAC=60° [ ALTERNATE INTERIOR ANGLES ARE EQUAL ]
In triangle ADC:-
∠ADC+∠DAC+∠DCA=180° [ ANGLE SUM PROPERTY OF A TRIANGLE ]
60°+60°+∠DCA=180°
120°+∠DCA=180°
∠DCA=180°-120°
∠DCA=60°
So, ∠DCA=60°
∠BAC=60°
∠ADC=60°
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