Question

∆ABC ~ ∆PQR for the correspondence ABC<-->PQR. 2m<P = 3m<Q and. m<C = 100. Find the m<b.

Note:- This sign before Alphabet means the angle (<).

1 one plz very imp ​

Answer 1

Answer:

32°

Step-by-step explanation:

Given

$\triangle ABC \sim \triangle PQR \: with \: correspondence \: to \\ABC \leftrightarrow PQR\\\\2\measuredangle P = 3\measuredangle Q \sf and, \: \measuredangle C = 100$

Since,

$\angle C \leftrightarrow \angle R,\\\\\measuredangle R = 100^\circ$

Now,

$2 \measuredangle P = 3 \measuredangle Q\\\\\implies \measuredangle P = \measuredangle \frac{3}{2} Q$

Now, by angle sum property,

$\angle P + \angle Q + \angle R = 180^\circ\\\\\implies \measuredangle \frac{3}{2} Q + \measuredangle Q + 100^\circ = 180^\circ\\\\(\rm since, \measuredangle P = \measuredangle \frac{3}{2} Q)\\\\ \implies\measuredangle \frac{3}{2} Q + \measuredangle \frac{2}{2}Q = 180^\circ - 100^\circ \\\\\implies \measuredangle \frac{5}{2} Q = 80^\circ\\\\\implies \measuredangle Q = 80 \times \frac{2}{5}\\\\\implies \measuredangle Q = 32^\circ$

Now, since

$\triangle ABC \sim \triangle PQR \\\textrm{with a correspondence of}\\\\ABC \leftrightarrow PQR\\\\\implies \measuredangle Q = \measuredangle B$

Hence, $\angle B = 32^\circ$

Answer 2

this is what you want it's this