Question

In △ABC the angle bisectors drawn from vertices A and B intersect at point
d. Find ∠ADB if∠A=50°, ∠B=100°

Answer 1

The measure of ∠ADB will be 105°

Explanation

According to the below diagram, the angle bisectors of ∠A and ∠B are intersecting at point D.

Given that, the measures of $\angle BAC = 50$° and $\angle ABC=100$°

So, $\angle DAB= \frac{50}{2}=25$° and $\angle DBA= \frac{100}{2}= 50$°

Now in triangle ADB, the sum of all three angles is 180°

So,  $\angle ADB= 180-(\angle DAB + \angle DBA)= 180-(25+50)=180-75= 105$

Thus, the measure of ∠ADB will be 105°

Answer 2

The measurement of angle D in the triangle ADB is: $\angle ADB = 105^\circ$

What are angle bisectors?

Angle bisectors are lines who bisect the considered angle. Bisect refers to splitting in two equal parts. Therefore, the bisected parts of the considered angle are half of the original angle.

Since the sum of angles in any triangle evaluates to $180^\circ$

and since the point D is the intersection of angle bisectors of angle A and B, therefore, in the triangle ADB, we have:
$m\angle BAD = m\angle A/2 = 50/2 = 25^\circ\\m\angle ABD = m\angle B/2 = 100/2 = 50^\circ$

Now, as sum of angles of a triangle is $180^\circ$, thus,

$m\angle ABD + m\angle BAD + m\angle ADB = 180^\circ\\25 + 50 + m\angle ADB = 180\\m\angle ADB = 180-75\\m\angle ADB = 105^\circ$

(remember that ∠ADB means angle between AD and DB, mainly that angle which is inside triangle ADB)

Thus, the measurement of angle D in the triangle ADB is: $\angle ADB = 105^\circ$

Learn more about angle bisectors here:

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