48. In the given figure, AB = BC, angle ABC = 68° DA & DB are the tangents to the circle with entre 0.
Calculate the measure of
(1) Angle ACB (ii) Angle AOB (1) Angle ADB
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Answers
Answer:
112
Step-by-step explanation:
Our aim is to calculate the measure of the asked angles.
We know that ∠ A B C = 68º and AB = BC.
We can say that in a triangle the sum of the three interior angles is equal to 180º.
Hence we have the following:
∠ C A B + ∠ A B C + ∠ A C B = 180 º
Opposite angles which are opposite to equal sides are equal. Then we can say that ∠ C A B = ∠ A C B.
Thus,
∠ A C B + ∠ A C B + 68 º = 180º
⇒ 2 ∠ A C B + 68º = 180 º
⇒ 2 ∠ A C B = 180º − 68 º
⇒ 2 ∠ A C B = 112 º
⇒ ∠ A C B = 112º / 2 = 56º
Hence, Angle ACB is equal to 56º
Since, a center angle is equal to the double of the angle at the circumference, we have that ∠ A O B = 2∠ A C B.
Then, our angle ∠ A O B is equal to 112º.
Hence, ∠ A O B = 112º
In order to compute the other ones, you just need to follow the same steps. Trivial
I hope this helps your studies! Keep it up!
Answer :
It is given that ∠ABC=68
and AB=BC.
In △ABC, the sum of three angles is 180
that is:
∠CAB+∠ABC+∠ACB=180
But ∠CAB=∠ACB (Angle opposite to equal sides are equal)
Therefore,
∠ACB+∠ACB+68
=180
⇒2∠ACB+68
=180
⇒2∠ACB=180
−68
⇒2∠ACB=112
⇒∠ACB=
112 /2
=56
Thus, ∠ACB=56
.
Now, ∠AOB=2∠ACB (Angle at centre is twice the angle at the circumference)
Therefore,
∠AOB=2×56
=112
Also, we know that ∠AOB+∠ADB=180
, thus
∠AOB+∠ADB=180
⇒112 +∠ADB=180
⇒∠ADB=180 −112
=68
Hence, ∠ADB=68
.