Question

$\huge \bf Maths \: Question$
ABC is a right triangle with the size of angle ACB equal to 74 degrees. The lengths of the sides AM, MQ and QP are all equal. Find the measure of angle QPB.

Note :- Refer to attachment for diagram
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Answer 1

IN TRIANGLE ABC,

BY ANGLE SUM PROPERTY 90+74+x=180

180-164=ANGLE QAM

ANGLE QAM=16

BUT AM=QM

SO, QAM=AQM

IN TRIANGLE AMQ,

BY ANGLE SUM PROPERTY 16+16+ANGLE AMQ=180

ANGLE AMQ=180-32

=148

BY LINEAR PAIR AMQ +QMP=180

148+QMP=180

QMP=32

BUT QM=QP

SO, QMP=QPM

BY LINEAR PAIR QPM+QPB=180

QPB=180-32

QPB=148

Answer 2

$\huge \bf {Question:-}$

In a given diagram,ABC is a right triangle with the measure of ∠ACB equal to 74°. The length of the sides AM,MQ and QP are all equal.Find the measure of ∠QPB.

$\huge \bf{Answer:-}$

Given:

• ∠ACB is measurement of 74°.
• ∠ABC is measurement of 90°.
• QM=QP=AM are equal length.

To find:

• measurement of ∠QPB.

Property used:

• Angle sum property of a triangle.
• Linear pair property.

According to Question:

In ∆ABC,

∠CAB+∠CBA+∠ACB=180° (Angle sum property of triangle)

∠CAB+90°+74°=180°

∠CAB=180°-90°-74°

∠CAB=16°. ....(i)

QM=QP ........(Given)

∠QMP=∠QPM ........(ii)

In ∆QPM,

∠QPM+∠QMP+∠MQP=180° (Angle sum property of triangle)

16°+16°+∠MQP=180°

∠MQP=180°-16°-16°

∠MQP=148°

Then,

∠AMQ+∠QMP=180° .......(By linear pair)

148°+∠QMP=180°

∠QMP=180°-148°

∠QMP=32°. .........(iii)

QMP=QPM ...................(by equation(ii))

∠QPM+∠QPB=180°

∠QPB+32°=180°

∠QPB=180°-32°

∠QPB=148°

Hence,∠QPB is measurement of 148°.

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