Question

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Question no17

Answer 1

Answer :

∆PQR ≈ ∆ABC

Scale factor = $\sf \frac{PQ}{AC} = \frac{8}{6}$

Steps of construction

1. Draw a line segment BC=5cm.

2. Construct OQ the perpendicular bisector of line segment BC meeting BC at P'.

3. Taking B and C as centre draw two arcs of equal radius 6 cm intersecting each other at A

Join BA and CA. So., is the required isosceles triangle

5. From, draw any ray BX making an acute .

6. Locate four points on BX such that .

Join and from draw a line intersecting the extended line segment BC at R.

7. From point R, draw meeting BA produced at P.

Then , ∆PQR Is required triangle.

Justification :

$\sf \frac{BC}{CR} =\frac{3}{1} \\ \sf \frac{BR}{BC} = \frac{BC+CR}{BC} \\ 1+ \frac{CR}{BC} = 1+ \frac{1}{3} = \frac{4}{3}$

also, RP parallel to CA

therefore,

∆ABC ≈ ∆ PBR

$\sf \frac{PB}{AB} = \frac{RP}{CA} = \frac{BR}{BC} = \frac{4}{3}$

Hence, the new triangle is similar to the given triangle.

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cosecθ + cotθ = m

cosecθ - cot θ= 1/cosecθ + cotθ =1/m

cosecθ + cot θ = m

cosecθ - cotθ =1/m/cosecθ =m+1/m/2

cotθ m-1/m/2

cosθ =m2-1/m2+1