Question

Pizzazz solve it correctly.

Answer 1

step by step explanation:

in triangle ABQ

tan 30° = AQ/ BQ

$\frac{1}{ \sqrt{3} } = \frac{3}{bq}$

then

$bq = 3 \sqrt{3} cm$

now in triangle PDC

sin45° = PD/CD

$\frac{1}{ \sqrt{2} } = \frac{3}{cd}$

$cd = 3 \sqrt{2} cm$

now in

triangle PDC

tan 45° = PD/PC

$bd = pc$

then pc = 3 cm

now BC = BQ + PC +QP

= (3√3 + 3 + 7)cm

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