# 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 = 2*x² – 2 x² *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* = 2*x² – 2 x² *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, 8*x² – 8* = 8x + 2*x²*

⇒ 6*x² – *8*x* *– 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**