Question

Peactice set 1.9
in the adjoining figure, beg po. seg QB. Beg RC
and Beg 3D osed to line lob = 6BC 9
CD: 12 and pg. 36, then find pe GR and
RS.
R.
ctives:
블
B​

Answer 1

Answer:

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Answer 2

Explanation:

and ∠C of ∆ABC intersect at point P. Prove that ∠BPC = 90° + 12∠BAC.

Complete the proof by filling in the blanks. (Textbook pg. no.27)

Maharashtra Board Class 9 Maths Solutions Chapter 3 Triangles Practice Set 3.1 13

Solution:

Proof:

In ∆ABC,

∠BAC + ∠ABC + ∠ACB = 180° …[Sum of the measures of the angles of a triangle is 180°]

∴ ∠BAC + – ∠ABC + ∠ACB = 180 … [Multiplying each term by 12]

∴ ∠BAC + ∠PBC + ∠PCB = 90°

∴ ∠PBC + ∠PCB = 90° – 1 ∠BAC ………(i)

In∆BPC,

∠BPC + ∠PBC + ∠PCB = 180° …….[Sum of measures of angles of a triangle]

∴ ∠BPC + 90° – 12∠BAC = 180° ……[From (i)]

∴ ∠BPC = 180° – 90°12∠BAC

= 180°- 90°+ 12∠BAC

= 90°+ 12∠BAC