Question

$\huge \mathfrak \red{Question}$
ABC is a right triangle. AM is perpendicular to BC. The size of angle ABC is equal to 55 degrees. Find the size of angle MAC. [Refer to the attachment for diagram]

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Answer 1

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°

Answer 2

Given :

• ABC is a right angle triangle.
• AM is perpendicular to BC. (AM ⊥ BC).
• ABC = 55°.

To find :

• The size of ∠MAC = ?

Concept used :

• Angle sum property.

Solution :

To find the ∠MAC = ?

By using angle sum property.

We need to find, ∠ACB = ?

$\implies \bf \angle ABC \ + \ \angle BAC \ + \ \angle ACB \ = \ 180^{\circ}$

$\implies \bf 55^{\circ} \ + \ 90^{\circ} \ + \ \angle ACB \ = \ 180^{\circ}$

$\implies \bf 145^{\circ} \ + \ \angle ACB \ = \ 180^{\circ}$

$\implies \bf \angle ACB \ = \ 180^{\circ} \ - \ 145^{\circ}$

$\qquad {\underline {\boxed {\tt \angle ACB \ = \ 35^{\circ}}}}$

Now,

Given that,

• ∠AMC = 90°.
• ∠ACM = 35°.

Now,

Finding the ∠MAC = ?

Using angle sum property,

$\implies \sf \angle AMC \ + \ \angle ACM \ + \ \angle MAC \ = \ 180^{\circ}$

$\implies \sf 90^{\circ} \ + \ 35^{\circ} \ + \ \angle MAC \ = \ 180^{\circ}$

$\implies \sf 125^{\circ} \ + \ \angle MAC \ = \ 180^{\circ}$

$\implies \sf \angle MAC \ = \ 180^{\circ} \ - \ 125^{\circ}$

$\qquad {\underline {\boxed {\tt \angle MAC \ = \ 55^{\circ}}}}$

•°• Therefore, The value of ∠MAC = 55°.