Selina solutions

Selina solutions

Grade 7

Triangles | Exercise 15(B)

Question 10

Q10) In the given figure, BI is the bisector of∠ABC and Cl is the bisector of ∠ACB. Find ∠BIC.

Solution:

n ∆ ABC,

BI is the bisector of ∠ABC and CI is the bisector of ∠ACB.

∵ AB = AC

∴ ∠B = ∠C

(Angles opposite to equal sides)

But ∠A = 40°

and ∠A + ∠B + ∠C = 180°

(Angles of a triangle)

⇒ 40° + ∠B + ∠B = 180°

⇒ 40° + 2∠B = 180°

⇒ 2∠B = 180° - 40° = 140°

⇒ ∠B = 140°/2 = 70°

∴ ∠ABC = ∠ACB = 70°

But BI and Cl are the bisectors of ∠ABC and ∠ACB respectively.

∠IBC = 1/2 ∠ABC = 1/2 (70°) = 35

and ∠ICB = 1/2 ∠ACB = 1/2 × 70°= 35

Now in ∆ IBC,

⇒ ∠BIC + ∠IBC + ∠ICB = 180°

(Angles of a triangle)

⇒ ∠BIC + 35° + 35° = 180°

⇒ ∠BIC = 180° - 70° = 110°

Hence ∠BIC = 110°

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subject-cta

Question 10

Q10) In the given figure, BI is the bisector of∠ABC and Cl is the bisector of ∠ACB. Find ∠BIC.

Solution:

n ∆ ABC,

BI is the bisector of ∠ABC and CI is the bisector of ∠ACB.

∵ AB = AC

∴ ∠B = ∠C

(Angles opposite to equal sides)

But ∠A = 40°

and ∠A + ∠B + ∠C = 180°

(Angles of a triangle)

⇒ 40° + ∠B + ∠B = 180°

⇒ 40° + 2∠B = 180°

⇒ 2∠B = 180° - 40° = 140°

⇒ ∠B = 140°/2 = 70°

∴ ∠ABC = ∠ACB = 70°

But BI and Cl are the bisectors of ∠ABC and ∠ACB respectively.

∠IBC = 1/2 ∠ABC = 1/2 (70°) = 35

and ∠ICB = 1/2 ∠ACB = 1/2 × 70°= 35

Now in ∆ IBC,

⇒ ∠BIC + ∠IBC + ∠ICB = 180°

(Angles of a triangle)

⇒ ∠BIC + 35° + 35° = 180°

⇒ ∠BIC = 180° - 70° = 110°

Hence ∠BIC = 110°

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Book a free class now

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subject-cta
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