Geometry Math Problem: Find the Height of the Triangle

Find the height of the triangle if

ABC is a right angle triangle, Where AB = 8 cm and PN = 1 cm

AB is the diameter of the semicircle and PN is the radius of the circle

AC & AB are tangents of the circle also AC is the tangent of the semicircle

Solution: Height of the triangle

We can redraw the diagram

Let MC = NC = x and ∠MCN = θ

Then heght of the triangle is 4 + x

From figure

∠PMC = ∠PNC = 90° (AC & BC are tangents)

⇒ ∠MPN = 360 – (∠PMC + ∠PNC + θ)

so, ∠MPN = 360 – (90 + 90 + θ)

⇒ ∠MPN = 180 – θ

AQ = BQ = 4 cm (radius of the semicircle)

PN = PM = 1 cm

AR = PM = 1 cm

⇒ RQ = 4 – 1 = 3 cm

From △PQR

PQ² = PR² + QR² (Pythagorean theorem)

⇒ 5² = PR² + 3²

then, PR = 4 cm

⇒ AM = PR = 4 cm

Apply cosine rule in triangle MCN

MN² = CM² + CN² – 2 × CM × CN × cos θ (Cosine Rule)

⇒ MN² = x² + x² – 2 × x × x × cos θ

⇒ MN² = 2x² – 2x² cos θ

Apply cosine rule in triangle MPN

MN² = PM² + PN² – 2 × PM × PN × cos (180-θ) (Cosine Rule)

⇒ MN² = 1 + 1+ 2 × 1 × 1 × cos θ

⇒ MN² = 2+ 2 cos θ

so, MN² = 2+ 2 cos θ = 2x² – 2x² cos θ

⇒ x² cos θ + cos θ = x² – 1

Now, cos θ = ( x² – 1 )/( x² + 1 )

⇒ tan θ = 2x/(x² – 1)

From △ABC

tan θ = AB/AC = AB/(AM + x)

⇒ tan θ = 8/(4 +x )

so, tan θ = 8/(4 +x) = 2x/(x² – 1)

⇒ 8( x² – 1 ) = 2x(4 +x)

so, 8x² – 8 = 8x + 2x²

⇒ 6x² – 8x – 8 = 0

This is a quadratic equation so apply quadratic equation formula

x=\frac{-8 \pm \sqrt{64 + 4 \times 6 \times 8}}{2 \times 6}

⇒ x = 2, -2/3

so we get x = 2 because length is always positive

AC = 4 + x = 4 + 2

⇒ Height of the triangle = 6 cm