Question

In the given figure, sides QP and RQ of ΔPQR are produced to points S and T respectively. If ∠SPR = 135º and ∠PQT = 110º, find ∠PRQ.



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Answer 1

Answer:

• ∠PRQ = 65°

Step-by-step explanation:

Given:-

• ∠SPR = 135°
• ∠PQT = 110°

To Prove:-

• ∠PRQ

Solution:-

It is given that,

∠SPR = 135º and ∠PQT = 110º

According to the question:-

⇒ ∠SPR + ∠QPR = 180º (Linear pair angles)

⇒ 135º + ∠QPR = 180º

⇒ ∠QPR = 45º

Also, ∠PQT + ∠PQR = 180º (Linear pair angles)

⇒ 110º + ∠PQR = 180º

⇒ ∠PQR = 70º

As the sum of all interior angles of a triangle is 180º, therefore, for ΔPQR,

⇒ ∠QPR + ∠PQR + ∠PRQ = 180º

⇒ 45º + 70º + ∠PRQ = 180º

⇒ ∠PRQ = 180º − 115º

⇒ ∠PRQ = 65º

Hence,

• ∠PRQ = 65º

Answer 2

Given

• ∠SPR = 135°
• ∠PQT = 110°

⠀

To find

• ∠PRQ

⠀

Solution

• As we can see in the given figure,

⠀

⠀⠀❍ ∠PQT and ∠PQR are forming linear ⠀⠀⠀⠀pair

⠀⠀

$\tt\longmapsto{∠PQT + ∠PQR = 180°}$

⠀

$\tt\longmapsto{110° + ∠PQR = 180°}\: \: \: \: \: \: \: {\bf{\bigg\lgroup{Given}{\bigg\rgroup}}}$

⠀

$\tt\longmapsto{∠PQR = 180° - 110°}$

⠀

$\tt\longmapsto{∠PQR = 70°}$⠀⠀...[1]

⠀

Similarly,

⠀

⠀⠀❍ ∠SPR and ∠QPR are forming linear ⠀⠀⠀⠀pair

⠀

$\tt\longmapsto{∠SPR + ∠QPR = 180°}$

⠀

$\tt\longmapsto{135° + ∠QPR = 180°}\: \: \: \: \: \: \: {\bf{\bigg\lgroup{Given}{\bigg\rgroup}}}$

⠀

$\tt\longmapsto{∠QPR = 180° - 135°}$

⠀

$\tt\longmapsto{∠QPR = 45°}$⠀⠀....[2]

⠀

We know that

⠀

$\boxed{\tt{\bigstar{Sum\: of\: all\: angles\: in\: a\: triangle = 180°{\bigstar}}}}$

⠀

$\tt\longrightarrow{∠PQR + ∠QPR + ∠PRQ = 180°}$

⠀

$\tt\longrightarrow{70° + 45° + ∠PRQ = 180°}\: \: \: \: \: \: \: {\bf{\bigg\lgroup{From\: 1\: and\: 2}{\bigg\rgroup}}}$

⠀

$\tt\longrightarrow{115° + ∠PRQ = 180°}$

⠀

$\tt\longrightarrow{∠PRQ = 180° - 115°}$

⠀

$\bf\longrightarrow{\boxed{\red{∠PRQ = 65°}}}$

⠀

⠀⠀❍ Hence

$\large{\boxed{\boxed{\sf{∠PRQ = 65°}}}}$

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