Question

In the given figure, AB = AC, AD ⊥ BC, ∠PAC = 102°. Find the values of x and y.

Answer 1

Given:

△ABC is isosceles with AB = AC

AD ⊥ BC → AD is the altitude from A to BC

∠PAC = 102°

We are asked to find x = ∠BAD and y = ∠ABC

Step 1: Analyze the figure

Since AB = AC, △ABC is isosceles with base BC

AD is altitude and in an isosceles triangle, the altitude from the vertex also bisects the base and the opposite angle.

So, BD = DC. 

∠BAD = ∠CAD

Let ∠BAD = x. 

Then, ∠CAD = x

Given ∠PAC = 102° → This is an exterior angle along AD.

Actually, ∠PAC = ∠BAD + ∠CAD = x + x = 2x or possibly 180 – 2x, depending on configuration.

Step 2: Relate given angle to x

AD is altitude → right angle at D

∠PAC = 102° is the angle at vertex A, outside the small triangle formed by the altitude

For an isosceles triangle with vertex A and base BC: ∠BAC = 2x

Since ∠PAC = 102°, it must be related to the exterior angle at A:

∠PAC = 180 – ∠BAC = 180 – 2x

Substitute the given:

180 – 2x = 102

2x = 78

x = 39°

Step 3: Reconsider configuration

∠PAC = ∠BAD + ∠DAC

In isosceles triangle,

∠BAD = ∠DAC = x

So, ∠PAC = ∠BAD + ∠DAC

= x + x

= 2x

Given ∠PAC = 102°

⇒ 2x = 102

⇒ x = 51°

Step 4: Find ∠ABC = y

In isosceles triangle, ∠ABC = ∠ACB = y

Sum of angles in triangle:

∠ABC + ∠ACB + ∠BAC = 180

y + y + ∠BAC = 180

2y + ∠BAC = 180

∠BAC = 2x = 102°

2y + 102 = 180

2y = 78

y = 39°