Question

In Fig. 6.39, sides QP and RQ of ΔPQR are produced to points S and T respectively. If ∠SPR = 135° and ∠PQT = 110°, find ∠PRQ.
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Answer 1

Answer:-

• ∠PRQ = 65°

Given:-

• ∠SPR = 135°
• ∠PQT = 110°

To Find:-

• ∠PRQ

Solution:-

ΔPQR are produced to points S and T respectively.

According to the question,

⇒ ∠SPR +∠QPR = 180° (SQ is a straight line.)

⇒ 135° +∠QPR = 180°

⇒ ∠QPR = 45°

also,

∠PQT +∠PQR = 180° (TR is a straight line.)

⇒ 110° +∠PQR = 180°

⇒ ∠PQR = 70°

Now,

⇒ ∠PQR +∠QPR + ∠PRQ = 180° (Sum of the interior angles of the triangle.)

⇒ 70° + 45° + ∠PRQ = 180°

⇒ 115° + ∠PRQ = 180°

⇒ ∠PRQ = 65°

Hence,

• ∠PRQ = 65°

Answer 2

Question:

In Fig. 6.39, sides QP and RQ of ΔPQR are produced to points S and T respectively. If ∠SPR = 135° and ∠PQT = 110°, find ∠PRQ.??

Given:

$\tt \red{ \angle PQT \: \angle SPR}$

Find:

$\tt \pink{PRQ}$

Solution:

$\sf \blue{According To The Question}$

Firstly,

$\orange\longrightarrow \sf \green{ \angle SPR \: + \:\angle QPR \longrightarrow 180 \degree(L.P.)} \\ \orange \longrightarrow \sf \green{135\degree +\angle QPR\longrightarrow180\degree} \\ \orange\longrightarrow\sf \green{\angle QPR \longrightarrow 180 \degree + 135 \degree} \\ \orange \longrightarrow \sf \green{\angle QPR\longrightarrow45 \degree}$

Now,

$\blue{\longrightarrow} \sf \purple{ \angle PQT + \angle PQR\longrightarrow 180\degree(L.P.)} \\ \blue{\longrightarrow} \sf \purple{110\degree + \angle PQR \longrightarrow180\degree} \\ \blue{\longrightarrow} \sf \purple{ \angle PQR\longrightarrow180\degree - 110} \\ \blue{\longrightarrow} \sf \purple{\angle PQR \longrightarrow70\degree}$

Now,

$\red{\longrightarrow} \sf \blue{sum \: of \: all \: angles \: of \:triangle \: is \: 180\degree} \\ \red{\longrightarrow} \sf \blue{ 45 \degree + 70\degree + x\longrightarrow180\degree(sum \: of \: all \: angles \: f \: triangle)} \\ \red{\longrightarrow} \sf \blue{115\degree + x\longrightarrow180\degree} \\ \red{\longrightarrow} \sf \blue{x\longrightarrow180\degree - 115\degree} \\ \red{\longrightarrow} \sf \blue{x\longrightarrow65\degree}$

Final Answer:

$\green \longrightarrow\tt \orange{x\longrightarrow65\degree}$