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In ΔABC the angle bisectors drawn from vertices A and B intersect at the point D. Find m∠ADB if m∠C = γ

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Answer:

The angle bisectors of a triangle intersect at a point called the incenter. In this case, the angle bisectors drawn from vertices A and B intersect at point D.

To find the measure of angle ADB (m∠ADB), we need to use the fact that the angle bisectors of a triangle divide the opposite side into segments that are proportional to the adjacent sides.

Let's assume that the length of AD is x and the length of BD is y. Since the angle bisectors divide the opposite side in a ratio equal to the ratio of the adjacent sides, we can set up the following proportion:

AC/BC = AD/BD

Now, since AC is opposite to angle C and BC is opposite to angle B, we can write the equation as:

sin(γ)/sin(∠C) = x/y

Simplifying this equation, we get:

y * sin(γ) = x * sin(∠C)

Now, to find m∠ADB, we need to find the value of x/y. We can do this by solving for x and y using the properties of triangles.

Using the Law of Sines, we know that in a triangle:

sin(∠C)/AC = sin(∠B)/BC

Since AC is opposite to angle C and BC is opposite to angle B, we can rewrite the equation as:

sin(∠C)/AD = sin(∠B)/BD

Simplifying this equation, we get:

BD * sin(∠C) = AD * sin(∠B)

Now, we can substitute this equation into the previous equation we derived:

y * sin(γ) = (AD * sin(∠B))

Simplifying further:

y = (AD * sin(∠B))/sin(γ)

Now, we have the values of x/y and y. We can substitute these values into the original equation:

sin(γ)/sin(∠C) = x/y

Substituting the values of x/y and y, we get:

sin(γ)/sin(∠C) = x/((AD * sin(∠B))/sin(γ))

Simplifying this equation, we get:

sin(γ)^2/sin(∠C) = x/(AD * sin(∠B))

Finally, to find the measure of angle ADB (m∠ADB), we can use the inverse sine function:

m∠ADB = arcsin(x/(AD * sin(∠B)))

This formula allows us to find the measure of angle ADB (m∠ADB) based on the given angle C (γ) and the properties of the triangle.

Explanation:

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