Question

If the decagon has all equal sides, and the measure of angle ATB is 120°, find the degree measure of angle TBP.

Answer 1

Consider the given decagon of all equal sides.

Given: $\angle ATB= 120^{\circ}$

To find: $\angle TBP= ?$

Construction: Join a dotted line from point T to P.

Now, $\angle ATB$ and $\angle BTP$ lies on the straight line. Hence, they form linear pair.

Therefore, $\angle ATB +\angle BTP= 180^{\circ}$

$120^{\circ} +\angle BTP= 180^{\circ}$

$\angle BTP= 180^{\circ}-120^{\circ}$

$\angle BTP = 60^{\circ}$ (equation 1)

Now, consider triangle TBP with two equal sides TB and BP.

By using the property

"Angles opposite to equal sides of an isosceles triangle are equal."

Since TB=BP, therefore the angles opposite to these sides are also equal.

$\angle BPT= \angle BTP$

From equation 1,

$\angle BPT= \angle BTP = 60^{\circ}$

Now, in triangle TBP,

By using angle sum property, which states

" The sum of measures of all three angles of a triangle is 180 degrees"

$\angle BPT+ \angle BTP+ \angle TBP= 180^{\circ}$

$60^{\circ}+60^{\circ}+ \angle TBP= 180^{\circ}$

$120^{\circ}+ \angle TBP= 180^{\circ}$

$\angle TBP= 180^{\circ}-120^{\circ}$

$\angle TBP= 60^{\circ}$

So, the measure of angle TBP is 60 degrees.