Question

The angle between the two altitudes of a parallelogram through the vertex of an obtuse angle of the parallelogram is 45°. Find the angles of the parallelogram.
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Answer 1

$\huge \red \star \: \: \huge \sf \underline{ \blue{given: }} \: \: \huge \red \star \:$

• ABCD is parallelogram.

• ∠EBF = 45°.

• BE and BF are altitudes on DC and DA respectively.

• ∠DEB = ∠BFA = 90°.

$\huge \red \star \: \: \huge \sf \underline{ \blue{to \: find : }} \: \: \huge \red \star \:$

• All angles of Parallelogram.

$\huge \red \star \: \: \huge \sf \underline{ \blue{solution: }} \: \: \huge \red \star \:$

In quadrilateral DEBF,

sum of all angles = 360°.

∠DEB + ∠EBF + ∠BFD +∠D = 360°

➥ 90 + 45 + 90 + ∠D = 360°

➥ 225 + ∠D = 360°

➥ ∠D = 135°

▬▬▬▬▬▬▬▬▬

∵ ABCD is parallelogram,

• Opposite angles are equal.

• Adjecent angles are supplementary (sum is 180°)

∴ ∠D = ∠B = 135°

➥ ∠D + ∠x = 180°

➥ 135 + ∠x = 180°

➥ ∠x = 45°

➙ All angles are 135°,135°,45°,45°.

Answer 2

Given:-

✯$\angle$ EBF = 45°

✯$\angle$ BFD = 90°

✯$\angle$ BED = 90°

Find:-

◕All angles of Paralleogram.

Diagram:-

$\setlength{\unitlength}{1 cm}\begin{picture}(0,0)\thicklines\qbezier(1,1)(1,1)(6,1)\qbezier(1,1)(1,1)(1.6,4)\qbezier(1.6,4)(1.6,4)(6.6,4)\qbezier(6,1)(6,1)(6.6,4)\qbezier(6,1)(1.2,2)(1.2,2)\qbezier(6,1)(5.5,4)(5.5,4)\qbezier(5.9,1.5)(5,1.9)(5.4,1.1) \qbezier(1.4,1)(1.6,1.4)(1.1,1.3)\qbezier(6,4)(5.9,3.2)(6.5,3.4) \qbezier(6,1)(1.2,2)(1.2,2) \qbezier(1.5,1.75)(1.2,1.8)(1.2,1.8)\qbezier(1.5,1.75)(1.6,2)(1.6,1.9)\put(4.7,1.4){ \bf 45^{\circ} } \put(5.7,3.2){ \bf x^{\circ} }\put(1.5,1.1){\bf x}\put(0.9,0.7){\bf A}\put(6,0.7){\bf B} \put(6.8,3.8){\bf C}\put(1.2,3.8){\bf D}\put(5.4,4.1){\bf E}\put(0.9,2){\bf F} \end{picture}$

Solution:-

In BEFD

$\sf \implies \angle BED + \angle FBE + \angle EDF + \angle DFB = 360^ \circ \bigg\lgroup Angle \: Sum \: Property\bigg\rgroup$

$\green{\sf where} \footnotesize{\begin{cases} \pink{\sf \angle BED = {90}^{ \circ}} \\ \purple{\sf \angle FBE = {45}^{ \circ}} \\ \orange{\sf \angle DFB = {90}^{ \circ}} \end{cases}}$

♡Substituting these values in the formula

$\sf \dashrightarrow \angle BED + \angle FBE + \angle EDF + \angle DFB = 360^ \circ$

$\sf \dashrightarrow {90}^{ \circ} + {45}^{ \circ} + \angle EDF + {90}^{ \circ} = 360^ \circ$

$\sf \dashrightarrow {225}^{ \circ} + \angle EDF= 360^ \circ$

$\sf \dashrightarrow \angle EDF= {135}^{ \circ}$

➠∠EDF = CBA = 135° 【Opp. angles of Paralleogram】

➲∠CDA + ∠DCB + ∠DAB + ∠CBA = 360°

〘Angle Sum Property〙

where,

• ∠CDA = 135°
• ∠CBA = 135°
• ∠DAB = ∠DCB = x

✧Substituting these values in formula:-

➲∠CDA + ∠DCB + ∠DAB + ∠CBA = 360°

➲135° + x + x + 135° = 360°

➲270° + 2x = 360°

➲2x = 360° - 270°

➲2x = 90°

➲x = 90/2

➲x = 45°

├┬┴┬┴┬┴┤├┬┴┬┴┬┴┤├┬┴┬┴┬┴┤├┬┴┬┴┬┴┤├┬┴┬┴┬┴┤

Hence, Angles of Paralleogram are:-

• ∠CDA = 135°
• ∠CBA = 135°
• ∠DAB = 45°
• ∠DCB = 45°