Question

Construct a ∆ ABC, in which
ZB = 60° ZC = 45° and AB + BC + CA = 11 cm.

Give justification also
━━━━━━━━━━━━━━━━━​

Answer 1

$\mathbb\red{Given, in \: ABC,}$

$\begin{gathered} \rm \footnotesize{ \implies \: \angle \: B \: = 60 \degree} \\ \rm \footnotesize{ \implies \: \angle C \: = 45 \degree}\end{gathered}$

$\rm \footnotesize{ ⇝ \: AB + BC + CA = 11 \: cm}$

$\tt\red{ STEPS \: \: OF \: \: CONSTRUCTION}$

$\bf \footnotesize \bold{(step - 1)}\rm \footnotesize{Draw \: a \: line \: segment \: PQ = 11 cm}$

$\begin{gathered} \rm \footnotesize \bold{(step - 2)}\rm \footnotesize{Draw \: ray \: PL \: making \: an \: \angle \: LPQ \: 60 \degree at \: P} \\ \rm \footnotesize{and \: ray \: QM \: making \: an \: \angle \: MQP = 45 \degree \: at \: Q}\end{gathered}$

$\begin{gathered} \rm \footnotesize \bold{(step - 3)}\rm \footnotesize{Bisect \: the \: \angle \: LPQ \: and \: \angle \: MQP, \: let \: the} \\ \rm \footnotesize{intersection \: point \: of \: these \: Bisectors \: be \: A \: }\end{gathered}</p><p>$

$\rm \footnotesize \bold{(step - 4)}\rm \footnotesize{Draw \: the \: \bot \: of \: ED \: of \: AP \: intersect \: PQ \: at \: \: B}(step−4)$

$\rm \footnotesize \bold{(step - 5)}\rm \footnotesize{Draw \: another \: \bot \: FG \: of \: AQ \: which \: intersect \: PQ \: at \: C}(step−5)$

$\rm \footnotesize \bold{(step - 6)}\rm$

Sɪɴᴄᴇ, B ʟɪᴇs ᴏɴ ᴛʜᴇ ᴘᴇʀᴘᴇɴᴅɪᴄᴜʟᴀʀ Bɪsᴇᴄᴛᴏʀ ED ᴏғ AP.

$\rm \small{so \: AB = BP \: \: then, \angle \: ABP = \angle \: PAB}$

Sɪᴍɪʟᴀʀʟʏ, C ʟɪᴇs ᴏɴ ᴛʜᴇ ᴘᴇʀᴘᴇɴᴅɪᴄᴜʟᴀʀ Bɪsᴇᴄᴛᴏʀ FG ᴏғ AQ.

$\rm \small{ \therefore \: AC = CQ \: then, \: \angle \: AQC = \angle \: QAC}$

$\rm \small{ Now,. BC + CA + AB = BC + CQ + BP = PQ}$

$\rm \small{ \angle \: ABC = \angle \: APB + \angle \: PAB = 2( \angle \: APB)}$

$\rm \small{ \implies \: \angle \: LPQ = 60 \degree}$

$\rm \small{Similarly \: \angle ACB = \angle \: AQC + \angle QAC}$

$\rm \small{ \implies \: 2( \angle \: AQC)}$

$\rm \small{ \implies \: \angle \: MQP = 45 \degree}$

Tʜᴜs, ᴛʜᴇ ᴄᴏɴsᴛʀᴜᴄᴛɪᴏɴ ɪs ᴊᴜsᴛɪғɪᴇᴅ...!!