Use the properties of the following special triangle to determine a.
We know that:
\widehat{A} + \widehat{B}+ \widehat{C} = 180^\circ
30^\circ + 75^\circ+ \widehat{C} = 180^\circ
\widehat{C} = 75^\circ
Therefore:
\widehat{B} = \widehat{C}
This is an isosceles triangle and we have:
AC = AB = 5
a=5
Use the properties of the following special triangle to determine a.
We have:
\widehat{B} = \widehat{C}
This is an isosceles triangle so point H is the midpoint of BC. Therefore:
BH=2
Using the Pythagoras theorem, we find:
a=\sqrt{5^2+2^2}= \sqrt{29}
a= \sqrt{29}
Use the properties of the following special triangle to determine a.
We have:
AC = AB
This is an isosceles triangle and:
\widehat{B} = \widehat{C}
We know that:
\widehat{A} + \widehat{B}+ \widehat{C} = 180^\circ
80^\circ+ 2\widehat{B} = 180^\circ
\widehat{B} = 50^\circ
\widehat{a}=50^\circ
Use the properties of the following special triangle to determine a.
We have:
AC = AB
This is an isosceles triangle and:
\widehat{B} = \widehat{C}
We know that:
\widehat{A} + \widehat{B}+ \widehat{C} = 180^\circ
60^\circ + \widehat{2 C} = 180^\circ
\widehat{C} = 60^\circ
This is also an equilateral triangle. So:
BC = 4
a=4
Use the properties of the following special triangle to determine a.
Using the Pythagoras theorem, we find:
\left(BA\right)^2=\left(AH\right)^2+\left(HB\right)^2
\left(AH\right)^2=\left({BA}\right)^2-\left(BH\right)^2
\left(AH\right)^2=\sqrt{80}^2-4^2 =64
Using the Pythagoras theorem, we find:
\left(AC\right)^2=\left(AH\right)^2+\left(HC\right)^2
\left(HC\right)^2=\left({AC}\right)^2-\left(AH\right)^2
\left(HC\right)^2=\sqrt{320}^2-64 =256
Therefore:
HC=16
a=16
Use the properties of the following special triangle to determine a.
We have:
AC = AB
This is an isosceles triangle and:
\widehat{B} = \widehat{C}
We know that:
\widehat{A} + \widehat{B}+ \widehat{C} = 180^\circ
50^\circ +2 \widehat{C} = 180^\circ
\widehat{C} = 65^\circ
\widehat{a} = 65^\circ
Use the properties of the following special triangle to determine a.
We know that:
\widehat{A} + \widehat{B}+ \widehat{C} = 180^\circ
Since the triangle is isosceles, we have:
\widehat{B} = \widehat{C}
90^\circ + 2\widehat{C} = 180^\circ
\widehat{C} = 45^\circ
\widehat{a} = 45^\circ