Question

11. If triangle ABC is an isosceles triangle and triangle DBE is an equilateral triangle, find each missing

Answer 1

Answer:

∠1 = 43°, ∠2 = 17°, ∠3 = 120°, ∠4 = ∠5 = ∠6 =  60°, ∠7 = 17°, ∠8 = 120°, ∠9 = 43°.

Step-by-step explanation:

I found the figure for this problem.

It is provided that ΔABC is isosceles and ΔDBE is equilateral.

Isosceles Triangle:

According to the definition the length of two sides are equal and hence the two base angles are equal.

Equilateral Triangle:

All the sides are of equal length and can also be defined as equiangular  with all the angles of measure 60°

CALCULATION:

For any triangle the sum of interior angles is always 180°.

ΔABC: ∠1+ (∠2 + ∠5 + ∠7)+∠9= 180°,

where, (∠2 + ∠5 + ∠7) is the total angle ∠ABC

ΔABC is isosceles so,

∠1 = ∠9,

4x+3 = 9x-47,

50 = 5x ⇒ x=10,

Putting the value of x in ∠1 we get,

∠1 = 4*(10)+3 = 43°,

∠1 = 43°.

ΔABD: ∠1+ ∠2+∠3= 180°

Now in order to find ∠2, firstly we have to find ∠3 in  ΔABD.

According to the exterior angle property, the exterior angle of an equilateral triangle is always 120°. Since ∠3 is an exterior angle for the equilateral triangle ΔDBE therefore,

∠3 =  120°,

furthermore,

∠2 = 180° - ∠3 - ∠1,

⇒ ∠2 = 180° - 120° - 43°

∠2 = 17°.

ΔDBE: ∠4+ ∠5+∠6= 180°,

Also, ΔDBE is equilateral ⇒ ∠4 = ∠5 = ∠6 =  60°.

Now from ∠1+ (∠2 + ∠5 + ∠7)+∠9= 180°

we can find ∠7,

⇒ ∠1+ (∠2 + ∠5 + ∠7)+∠9= 180°

∠7 =  180° - (∠2 + ∠5 + ∠1+∠9),

∠7 =  180° - (17° + 60° + 43° + 43°),

∠7 =  180° - (163°)  

∠7 = 17°.

Similarly, ∠8 is an exterior angle for the equilateral triangle ΔDBE therefore,

∠8 =  120°,

Answer 2

Answer:

Step-by-step explanation: