Question

× Kindly solve the given question in the attachment. (step - by - step) ​

Answer 1

any pair of angles each of which is on the same side of one of two lines cut by a transversal and on the same side of the transversal.

Answer 2

$\begin{gathered}\bf{\underline{\underline{Given :-}}}\\\end{gathered}$

• AP ⊥ BC

• BQ ⊥ AC

$\begin{gathered}\bf{\underline{\underline{To\:Prove / To\:Find : - }}}\\\end{gathered}$

1. △CPA ~ △CQB.
2. Proportionality of both triangles.
3. Find AC.

$\begin{gathered}\bf{\underline{\underline{Solution :-}}}\\\end{gathered}$

• △CPA ~ △CQB.

In △CPA and △CQB.

∠P = ∠Q - Each 90°

∠C = ∠C - Common angle.

$\rm\blue{★ \ Therefore \ , △CPA ~ △CQB \ - By \ AA \ Similarity.}$

_________________________

• Proportionality of both triangles.

△CPA ~ △CQB

$\sf{CP / CQ = PA / QB = AC / BC - By \ C.P.C.T ---- (1)}$

_________________________

• Find AC.

PA / QB = AC / BC - From (1)

7 / 8 = AC / 12

$\rm\pink{★ \ AC \ = \ 10.5 \ units.}$