Answer :
Answer:
4.
∠ABC + ∠BCA + ∠CAB = 180
therefore, ∠ABC = 180 - ∠BCA - ∠CAB
using this formula to solve 4a, 4b and 4c:
4a. ∠ABC = 180 - ∠BCA - ∠CAB = 180 - 49 -57 = 26
∠ABC = 26
4b. ∠ABC = 180 - ∠BCA - ∠CAB = 180 - 28 -90 = 62
∠ABC = 62
4c. ∠ABC = 180 - ∠BCA - ∠CAB = 180 - 38 - 25 = 117
∠ABC = 117
5a. In the Triangle ABC, ∠BCA + ∠CAB + ∠ABC = 180
∠BCA = 180 - ∠CAB - ∠ABC = 180 - 55 - 60 = 65
∠BCA = 65
∠BCD + ∠BCA = 180
therefore substituting for ∠BCA in, ∠BCD = 180 - ∠BCA = 180 - 65 = 115
∠BCD =115
5b. In the Triangle ABC, ∠BCA + ∠CAB + ∠ABC = 180
∠BCA = 180 - ∠CAB - ∠ABC = 180 - 125 - 30 = 25
∠BCA = 25
∠BCD + ∠BCA = 180
therefore substituting for ∠BCA in, ∠BCD = 180 - ∠BCA = 180 - 25 = 155
∠BCD = 155
5c. ∠ABD + ∠CBD = 180
therefore ∠CBD = 180 - ∠ABD = 180 - 132 = 48
∠CBD = 48
In the Triangle BCD, ∠CBD + ∠BDC + ∠BCD = 180
therefore ∠BCD = 180 - ∠CBD - ∠BDC = 180 - 48 - 71 = 61
∠BCD =61
6. 4 angles of a quadrilateral = 360
∠a+110+60+80=360
∠a=360-110-60-80=110
∠a = 110
7. four angles of a quadrilateral = 360
using his formula to solve for ∠a, ∠b and ∠c.
for angle a: ∠a+110+95+63=360
∠a=360-110-95-63=92
∠a = 92
for angle b: ∠b+40+62+35=360
∠b=360-40-62-35=223
∠b=223
for angle c: ∠c+100+172+35=360
∠c=360-100-172-35=53
∠c=53