Answer:
2. <T=41°,<X=90°,<O=49
1. <B=41°,<X=49°,<O=90°
Explanation:
<E from ∆BEO=<E from ∆TEX
4x-2=3x+23
C.L.T(Collect like terms)
4x-3x=23+2
x=25
cross checking
4(25)-2=3(25)+23
100-2=75+23
98=98
so it's correct
so,in∆TEX
base angles of an isosceles are equal
so 2x+5 becomes X
98+X+X=180
98+2X=180
C.L.T
2X=180-98
2X=82
divide both sides by 2
X=41
so to find x in<T
<T=2x-5=41
2x=41+5
2x=46
divide both sides by 2
x=23
2. for ∆TXO
the following angles are given
<T=41,<X=90<O=x
<T+<X+<O=180
<O+41+90=180
<O+131=180
C.L.T
<O=180-131
<O=49
remember<X from ∆TEX=41
<X from ∆TOX=90
so 90-41
<X from ∆BOX=49°
1. for ∆BOX
<B+<O+<X=180
let angle B be y
y+90+49=180°
y+139=180
C.L.T
y=180-139
y=41°
reason why <B=<T
because Adjacent angles are equal