ABC is a right angled triangle with angle angle a=90 AM is perpendicular to BC and measure of angle ABC is 55 degrees . Then the measure of angle MAC isdegrees.
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Answered by
2
Answer:
As
A
D
is drawn perpendicular to
B
C
in right angled
Δ
A
B
C
, it is apparent that
Δ
A
B
C
is right angled at
∠
A
as shown below (not drawn to scale).
enter image source here
As can be seen
∠
B
is common in
Δ
A
B
C
as well as
Δ
D
B
A
(here we have written two triangles this way as
∠
A
=
∠
D
,
∠
B
=
∠
B
and
∠
C
=
∠
B
A
D
) - as both are right angled (obviously third angles too would be equal) and therefore we have
Δ
A
B
C
≈
Δ
D
B
A
and hence
B
C
A
B
=
A
B
B
D
=
A
C
A
D
..............(1)
therefore, we have
B
C
A
B
=
A
B
B
D
or
A
B
2
=
B
C
×
B
D
=
9
×
4
=
36
Hence
A
B
=
6
Answered by
2
Given:-
- ABC is a right-angled triangle at A
- AM is perpendicular to BC, i.e. AM ⊥ BC
- And ∠ABC = 55°
To find:-
- ∠MAC
In triangle ABC, we have,
- ∠ABC = 55°
- ∠BAC = 90°
By angle sum property if a triangle,
The sum of all three angles of a triangle is 180°.
- ⇒ ∠ABC + ∠BAC + ∠ACB = 180°
- ⇒ 55° + 90° + ∠ACB = 180°
- ⇒ 145° + ∠ACB = 180°
- ⇒ ∠ACB = 180° - 145°
- ⇒ ∠ACB = 35°
Now,
In triangle AMC, we have,
- ∠AMC = 90°
- ∠ACM = 35°
By angle sum property if a triangle,
The sum of all three angles of a triangle is 180°.
- ⇒ ∠AMC + ∠ACM + ∠MAC = 180°
- ⇒ 90° + 35° + ∠MAC = 180°
- ⇒ 125° + ∠MAC = 180°
- ⇒ ∠MAC = 180° - 125°
- ⇒ ∠MAC = 55°
Hence,
The value of ∠MAC is 55°
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