Question

In a triangle abc if ab=ac and bc is produced to d such that angle acd = 100 degrees, then angle a =

Answer 1

Given:

In Δ ABC,

AB = AC

BC is produced to D such that ∠ACD = 100°

To find:

The measure of ∠A

Solution:

It is given that,

AB = AC

∴ ∠ABC = ∠ACB ....... [Angles opposite to equal sides are also equal in length] ...... (i)

We have,

∠ACD + ∠ACB = 180° ..... [Linear Pair]

⇒ 100° + ∠ACB = 180°

⇒ ∠ACB = 180° - 100°

⇒ ∠ACB = 80° ..... (ii)

From (i) & (ii), we get

∠ABC = ∠ACB = 80°

Now,

∠ABC + ∠ACB + ∠A = 180° ..... [Sum of the 3 interior angles of a triangle is 180°]

substituting the values of ∠ABC & ∠ACB, we get

⇒ 80° + 80° + ∠A = 180°

⇒ 160° + ∠A = 180°

⇒ ∠A = 180° - 160°

⇒ ∠A = 20°

Thus, the measure of ∠A is 20°

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Answer 2

Step-by-step explanation:

ABC is a triangle where AB=BC and BC is extended to point D and AD is joined & BC=CD. Now, what is the value of the 3 angles of triangle ACD?

Answered September 3

Here, AB = BC = CD and B- C - D

So, BD = 2AB

In triangle ABC, angle BAC = angle ACB =90 - (B/2).

angle ACD is an exterior angle of triangle ABC, so angle ACD = B + 90 - (B/2)

= 90 + (B/2)

Suppose angle CAD = x, so angle D = 180 - x - (90 + (B/2)) = 90 - (B/2) - x

angle BAD = 90 - (B/2) + x

In triangle ABD, AB /sin D = BD / sin BAD

sin((90 - (B/2)) + x ) = 2 AB sin D/AB

= 2 sin D = 2 sin ((90 - (B/2)) - x )

So, cos ((B/2) - x) = 2 cos ((B/2) + x )

So, cos (B/2)*cos x + sin (B/2)*sin x = 2 cos(B/2)*cos x - 2 sin (B/2)* sin x

So, 3 sin (B/2)*sin x = cos (B/2)* cos x

So, tan x = (1/3)*cot (B/2)

The question has infinite solutions.

(1) If B = 90 degree then x = tan inverse (1/3)

(2) If B = 60 degree then x = 30 degree.

Thus, for different values of B, we can find the values of x and then values of other two angles of triangle ACD.