Question

in the glven figure, Br is the bisector of ZABC and Cl is the bisector of ZACB. Find ZBIC​

Answer 1

Answer:

${ \underline{ \boxed{\mathbb{\red{given}}}}} \\ In ∆ ABC, \\ </p><p></p><p>BI is the bisector of ∠ABC and CI is the bisector of ∠ACB. \\ </p><p></p><p></p><p></p><p>∵ AB = AC \\ </p><p></p><p>∴ ∠B = ∠C \\ </p><p></p><p>(Angles opposite to equal sides) \\ </p><p></p><p> But ∠A = 40° \\ </p><p></p><p> and ∠A +  ∠B + ∠C = 180° \\ </p><p></p><p>(Angles of a triangle) \\ </p><p></p><p>⇒ 40° + ∠B + ∠B = 180° \\ </p><p></p><p>⇒ 40° + 2∠B = 180° \\ </p><p></p><p>⇒ 2∠B = 180° - 40° = 140° \\ </p><p></p><p>⇒ ∠B = 140°/2 = 70° \\ </p><p></p><p>∴ ∠ABC = ∠ACB = 70° \\ </p><p></p><p>{ \underline{ \boxed{\mathbb{\red{But }}}}}BI and Cl are the bisectors of ∠ABC and ∠ACB respectively. \\ </p><p></p><p> ∠IBC = 1/2 ∠ABC = 1/2 (70°) = 35 \\ </p><p></p><p> and ∠ICB = 1/2 ∠ACB = 1/2 × 70°= 35 \\ </p><p></p><p>Now in ∆ IBC, \\ </p><p></p><p>⇒ ∠BIC + ∠IBC + ∠ICB = 180° \\ </p><p></p><p>(Angles of a triangle) \\ </p><p></p><p>⇒   ∠BIC + 35° + 35° = 180° \\ </p><p></p><p> ⇒ ∠BIC = 180° - 70° = 110° \\ </p><p></p><p>{ \underline{ \boxed{\mathbb{\pink{Hence }}}}} ∠BIC = 110° \\ </p><p></p><p>$

Step-by-step explanation:

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