Question

Find the measurement of the angles of triangle BXO and TXO

Answer 1

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