In parallelogram ABCD, ∆BCE = 65° and ∆ACD = 35º. DC is extended to E. Find the magnitude of:—
a. ∆ABC
b. ∆ADC
C. ∆ACB
Whoever he/she will give me the correct answer I will definately make him/her the Brainliest. That's my promise to you.
a. ∆ABC
b. ∆ADC
C. ∆ACB
Whoever he/she will give me the correct answer I will definately make him/her the Brainliest. That's my promise to you.
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Step-by-step explanation:
Let ABCD be a parallelogram in which ∠A=65
Let ABCD be a parallelogram in which ∠A=65 ∘
Let ABCD be a parallelogram in which ∠A=65 ∘ .
Let ABCD be a parallelogram in which ∠A=65 ∘ .Since AD∥BC we can treat AB as a transversal. So,
Let ABCD be a parallelogram in which ∠A=65 ∘ .Since AD∥BC we can treat AB as a transversal. So,∠A+∠B=180
Let ABCD be a parallelogram in which ∠A=65 ∘ .Since AD∥BC we can treat AB as a transversal. So,∠A+∠B=180 ∘
Let ABCD be a parallelogram in which ∠A=65 ∘ .Since AD∥BC we can treat AB as a transversal. So,∠A+∠B=180 ∘
Let ABCD be a parallelogram in which ∠A=65 ∘ .Since AD∥BC we can treat AB as a transversal. So,∠A+∠B=180 ∘ 65
Let ABCD be a parallelogram in which ∠A=65 ∘ .Since AD∥BC we can treat AB as a transversal. So,∠A+∠B=180 ∘ 65 ∘
Let ABCD be a parallelogram in which ∠A=65 ∘ .Since AD∥BC we can treat AB as a transversal. So,∠A+∠B=180 ∘ 65 ∘ +∠B=180
Let ABCD be a parallelogram in which ∠A=65 ∘ .Since AD∥BC we can treat AB as a transversal. So,∠A+∠B=180 ∘ 65 ∘ +∠B=180 ∘
Let ABCD be a parallelogram in which ∠A=65 ∘ .Since AD∥BC we can treat AB as a transversal. So,∠A+∠B=180 ∘ 65 ∘ +∠B=180 ∘
Let ABCD be a parallelogram in which ∠A=65 ∘ .Since AD∥BC we can treat AB as a transversal. So,∠A+∠B=180 ∘ 65 ∘ +∠B=180 ∘ ∠B=180
Let ABCD be a parallelogram in which ∠A=65 ∘ .Since AD∥BC we can treat AB as a transversal. So,∠A+∠B=180 ∘ 65 ∘ +∠B=180 ∘ ∠B=180 ∘
Let ABCD be a parallelogram in which ∠A=65 ∘ .Since AD∥BC we can treat AB as a transversal. So,∠A+∠B=180 ∘ 65 ∘ +∠B=180 ∘ ∠B=180 ∘ −65
Let ABCD be a parallelogram in which ∠A=65 ∘ .Since AD∥BC we can treat AB as a transversal. So,∠A+∠B=180 ∘ 65 ∘ +∠B=180 ∘ ∠B=180 ∘ −65 ∘
Let ABCD be a parallelogram in which ∠A=65 ∘ .Since AD∥BC we can treat AB as a transversal. So,∠A+∠B=180 ∘ 65 ∘ +∠B=180 ∘ ∠B=180 ∘ −65 ∘
Let ABCD be a parallelogram in which ∠A=65 ∘ .Since AD∥BC we can treat AB as a transversal. So,∠A+∠B=180 ∘ 65 ∘ +∠B=180 ∘ ∠B=180 ∘ −65 ∘ ∠B=115
Let ABCD be a parallelogram in which ∠A=65 ∘ .Since AD∥BC we can treat AB as a transversal. So,∠A+∠B=180 ∘ 65 ∘ +∠B=180 ∘ ∠B=180 ∘ −65 ∘ ∠B=115 ∘
Let ABCD be a parallelogram in which ∠A=65 ∘ .Since AD∥BC we can treat AB as a transversal. So,∠A+∠B=180 ∘ 65 ∘ +∠B=180 ∘ ∠B=180 ∘ −65 ∘ ∠B=115 ∘
Let ABCD be a parallelogram in which ∠A=65 ∘ .Since AD∥BC we can treat AB as a transversal. So,∠A+∠B=180 ∘ 65 ∘ +∠B=180 ∘ ∠B=180 ∘ −65 ∘ ∠B=115 ∘ Since the opposite angles of a parallelogram are equal, we have
Let ABCD be a parallelogram in which ∠A=65 ∘ .Since AD∥BC we can treat AB as a transversal. So,∠A+∠B=180 ∘ 65 ∘ +∠B=180 ∘ ∠B=180 ∘ −65 ∘ ∠B=115 ∘ Since the opposite angles of a parallelogram are equal, we have∠C=∠A=65
Let ABCD be a parallelogram in which ∠A=65 ∘ .Since AD∥BC we can treat AB as a transversal. So,∠A+∠B=180 ∘ 65 ∘ +∠B=180 ∘ ∠B=180 ∘ −65 ∘ ∠B=115 ∘ Since the opposite angles of a parallelogram are equal, we have∠C=∠A=65 ∘
Let ABCD be a parallelogram in which ∠A=65 ∘ .Since AD∥BC we can treat AB as a transversal. So,∠A+∠B=180 ∘ 65 ∘ +∠B=180 ∘ ∠B=180 ∘ −65 ∘ ∠B=115 ∘ Since the opposite angles of a parallelogram are equal, we have∠C=∠A=65 ∘ and ∠D=∠B=115
Let ABCD be a parallelogram in which ∠A=65 ∘ .Since AD∥BC we can treat AB as a transversal. So,∠A+∠B=180 ∘ 65 ∘ +∠B=180 ∘ ∠B=180 ∘ −65 ∘ ∠B=115 ∘ Since the opposite angles of a parallelogram are equal, we have∠C=∠A=65 ∘ and ∠D=∠B=115 ∘
Let ABCD be a parallelogram in which ∠A=65 ∘ .Since AD∥BC we can treat AB as a transversal. So,∠A+∠B=180 ∘ 65 ∘ +∠B=180 ∘ ∠B=180 ∘ −65 ∘ ∠B=115 ∘ Since the opposite angles of a parallelogram are equal, we have∠C=∠A=65 ∘ and ∠D=∠B=115 ∘
Let ABCD be a parallelogram in which ∠A=65 ∘ .Since AD∥BC we can treat AB as a transversal. So,∠A+∠B=180 ∘ 65 ∘ +∠B=180 ∘ ∠B=180 ∘ −65 ∘ ∠B=115 ∘ Since the opposite angles of a parallelogram are equal, we have∠C=∠A=65 ∘ and ∠D=∠B=115 ∘ Hence, ∠B=115
Let ABCD be a parallelogram in which ∠A=65 ∘ .Since AD∥BC we can treat AB as a transversal. So,∠A+∠B=180 ∘ 65 ∘ +∠B=180 ∘ ∠B=180 ∘ −65 ∘ ∠B=115 ∘ Since the opposite angles of a parallelogram are equal, we have∠C=∠A=65 ∘ and ∠D=∠B=115 ∘ Hence, ∠B=115 ∘
Let ABCD be a parallelogram in which ∠A=65 ∘ .Since AD∥BC we can treat AB as a transversal. So,∠A+∠B=180 ∘ 65 ∘ +∠B=180 ∘ ∠B=180 ∘ −65 ∘ ∠B=115 ∘ Since the opposite angles of a parallelogram are equal, we have∠C=∠A=65 ∘ and ∠D=∠B=115 ∘ Hence, ∠B=115 ∘ ,∠C=65
Let ABCD be a parallelogram in which ∠A=65 ∘ .Since AD∥BC we can treat AB as a transversal. So,∠A+∠B=180 ∘ 65 ∘ +∠B=180 ∘ ∠B=180 ∘ −65 ∘ ∠B=115 ∘ Since the opposite angles of a parallelogram are equal, we have∠C=∠A=65 ∘ and ∠D=∠B=115 ∘ Hence, ∠B=115 ∘ ,∠C=65 ∘
Let ABCD be a parallelogram in which ∠A=65 ∘ .Since AD∥BC we can treat AB as a transversal. So,∠A+∠B=180 ∘ 65 ∘ +∠B=180 ∘ ∠B=180 ∘ −65 ∘ ∠B=115 ∘ Since the opposite angles of a parallelogram are equal, we have∠C=∠A=65 ∘ and ∠D=∠B=115 ∘ Hence, ∠B=115 ∘ ,∠C=65 ∘ and ∠D=115