In δabc, ad and be are the angle bisectors of ∠a and ∠b and de ║ ab . find the measures of the angles of δabc, if m∠ade: m∠adb = 2: 9.

∠ A = ∠B= 48° and ∠ C = 84°

Step-by-step explanation:

Here, m∠ADE: m∠ADB = 2:9

Let m∠ADE = 2x and m∠ADB = 9x

Where x is any value.

By joining the points D and E (construction)

Since, Here DE ║ AB.

⇒ ∠DAB = 2 x

⇒ ∠ A = 4x ( because AD is the angle bisector so, ∠DAB=∠DAE = 2x )

Now,  Let O is the intersection point of angle bisectors AD and BE.

Then, By the property of angle bisctor.

O is the circumcenter of the triangle ABC.

Therefore, OA = OB

⇒ ∠DAB = ∠EBA = 2 x

But BE is the angle bisceor,

Therefore, ∠EBA = ∠EBC=2 x

But, ∠B = ∠EBA + ∠EBC

⇒ ∠B = 4x

Now, since BD is the same transversal on the parallel lines AB and ED,

⇒ ∠B = ∠ EDC

⇒ ∠ EDC = 4x

Since, ∠ADB + ∠ADE + ∠EDC = 180°

⇒ 9x + 2x + 4x = 180°

⇒ 15x = 180°

⇒ x = 12°

Thus, the measures of the angles of ΔABC are,

∠ A = 4x =4×12 = 48°

∠ B = 4x =4×12= 48°

⇒ ∠ C = 180° - 48°- 48°= 84°

step-by-step explanation: given equation a = h(b 1+b 2)

also we are given values of a, h and b 1 as :

a = 16, h = 4, and b 1 = 3.

in order to solve it for b 2, we need to plug the given values of   a, h and b 1 .

therefore, plugging values of   a, h and b 1   in given equation, we get

16 = 4( 3 + b 2)

first dividing both sides by 4, we get

4 = 3 + b 2

subtracting 3 from both sides, we get

4 -3 = 3 -3 + b 2

1 = b 2.