Question

Q. Find :-
(i) <ACB ? ​

Answer 1

Given,

angle A = 55°

angle CDB = 100°

angle DCB = 25°

find: angle ACB.

BCD is a triangle

hence, all the angles add up to 180° (interior angle property)

we know 2 angles, so the third angle will be 180-(sum of two angles) = 180-(100+25) = 180°-125° = 55°

angle B = 55°

ABC is a triangle

here, angle C = angle DCB + angle ECD (let angle ECD be x)

hence by angle sum property, in triangle ABC,

angle A+B+C = 180°

55°+55°+25°+x = 180°

x = 180° - (55°+55°+25°)

x = 180° - 135°

x =. 45°

hence angle ACB/C = 25°+45° = 70°. {as angle C = angle DCB+ECD}

hope it helps

have a great day

Answer 2

find :

<ACB ?

Step-by-step explanation:

given points:-

∆ABC ,

<DAE=55°

<DCB=25°

<CDB=100°

BCD=180° ...... (1) [ sum of trigle is 180° ]

< DBC+ <DCB + < CDB =180°

<B + <C + <D=180°

<B + 25° + 100° = 180°

<B + 125° = 180°

<B = 180°- 125°

<B = 55°

<DBC = 55° ..... (2)

∆ ABC = 180° ....(3)

<CAB + < ABC + < ACB = 180°

<A + <B + <C = 180°

55° + 55° + <ACB = 180° ... [ < B = 55° from (2) ]

110° + < ACB = 180°

< ACB = 180° - 110°

<ACB= 70°

FINAL ANSWER <ACB = 70° .