What are the shapes of the triangles?
(1) Given 12, 15, 20 be the sides of the triangle ABC . If we use
the altitudes ha, hb and hc of ABC as lengths to construct
another XYZ. What kind of triangle is XYZ ?
Method 1
1
1
s = 2 (ha + hb + hc ) = 2 (12 + 15 + 20) =
47
2
By Heron Formula, the area of XYZ , S is
47 47
47
47
2S
√128639
10
S = √s(s − ha )(s − hb )(s − hc ) = √ 2 ( 2 − 12) ( 2 − 15) ( 2 − 20) =
2S
ha = 12 =
Hence
2√128639
hb 2 + hc 2 = (
2
=
2
hb = 15 =
√128639
)
10
) +(
15
√128639
)
6
ha 2 = (
2S
√128639
,
6
2
=
2√128639
15
, hc = 20 =
√128639
4
,
128639
36
128639
36
∴ hb 2 + hc 2 = ha 2
By the Converse of Pythagoras Theorem, XYZ is a right angled triangle.
Obviously XYZ is scalene, since all three sides ha, hb and hc are different.
The calculation of the exact length of the altitudes may be difficult. A better method is shown
below.
Method 2
Let
S be the area of the triangle, then
1
1
1
1
1
1
S = 2 ha (12) = 2 hb (15) = 2 hc (20)
60 60 60
ha : hb : hc = 12 : 15 : 20 = 12 : 15 : 20 = 5: 4: 3 ⟹ ha = 5k, hb = 4k, hc = 3k
Since
hb 2 + hc 2 = (4k)2 + (3k)2 = 25k 2 = ha 2
By the Converse of Pythagoras Theorem, XYZ is a right angled triangle.
Obviously XYZ is scalene, since all three sides ha : hb : hc = 5: 4: 3 and are different.
(2) Given that a , b and c are the three sides of ABC,
If c 2 (a2 + b2 − c 2 ) = b2 (c 2 + a2 − b2 ) , then what kind of triangle is
ABC ?
c 2 (a2 + b2 − c 2 ) = b2 (c 2 + a2 − b2 )
c 2 a2 + b 2 c 2 − c 4 − b 2 c 2 − a2 b 2 + b 4 = 0
1
b4 − c 4 − a2 (b2 − c 2 ) = 0
(b2 + c 2 )(b2 − c 2 ) − a2 (b2 − c 2 ) = 0
(b2 − c 2 )(b2 + c 2 − a2 ) = 0
(i)
If
(ii) If
b2 − c 2 = 0,
b2 = c 2 ,
b2 + c 2 − a2 = 0 , then a2 = b2 + c 2 .
By the Converse of Pythagoras Theorem,
(3) In
ABC is isosceles.
b = c , then
ABC, if
cos A+2 cos C
cos A+2 cos B
ABC is a right angled triangle with
ABC = 90o .
sin B
= sin C , then what is the shape of the triangle ?
Method 1
Use Cosine Law and Sine Law for
b2 +c2 −a2
a2 +b2 −c2
+2
2bc
2ab
b2 +c2 −a2
a2 +c2 −b2
+2
2bc
2ac
cos A+2 cos C
cos A+2 cos B
=
sin B
sin C
b
=
c
a(b2 +c2 −a2 )+2c(a2 +b2 −c2 )
a(b2 +c2 −a2 )+2b(a2 +c2 −b2 )
=
b
c
ac(b2 + c 2 − a2 ) + 2c 2 (a2 + b2 − c 2 ) = ab(b2 + c 2 − a2 ) + 2b2 (a2 + c 2 − b2 )
a(b2 + c 2 − a2 )(c − b) + 2[a2 (c 2 − b2 ) − (c 4 − b4 )] = 0
(c − b)[a(b2 + c 2 − a2 ) + 2(c + b)(a2 − c 2 − b2 )] = 0
(c − b)(b2 + c 2 − a2 )(a − 2b − 2c) = 0
Since by Triangular Inequality, 2b + 2c > 2𝑎 > 𝑎 ⟹ 𝑎 − 2𝑏 − 2𝑐 ≠ 0.
∴ c = b or b2 + c 2 = a2
Therefore ABC is an isosceles triangle or a right-angled triangle with A as right- .
Method 2
cos A+2 cos C
cos A+2 cos B
=
sin B
sin C
cos A sin C + 2 sin C cos C = cos A sin B + 2 sin B cos B
cos A sin C + sin 2C = cos A sin B + sin 2B
cos A (sin C − sin B) + (sin 2C − sin 2B) = 0
cos A [2cos (
cos A [2cos (
C+B
C−B
2
2
) sin (
)] − 2cos (C + B) sin(C − B) = 0
180°−A
C−B
2
2
) sin (
)] − 2cos (180° − A) sin(C − B) = 0
2
A
C−B
C−B
C−B
2
2
2
cos A [2sin ( 2 ) sin (
C−B
2cos A sin (
)] + 2cos A [ 2sin (
A
C−B
cos A = 0 or sin (
(i)
2
2
2
) = 0 , then
A
2
A
2
)=0
= 0 , ∠C = ∠B
and
ABC is an isosceles triangle.
A
< 90° ⟹ sin ( 2 ) > 0
(b) −90° <
(c)
C−B
∠A = 90° , ABC is a right-angled triangle.
C−B
(iii) (a) 0° <
)] = 0
A
C−B
(ii) If sin (
2
) = 0 or sin ( 2 ) + 2 cos (
If cos A = 0, then
)] = 0
C−B
) [sin ( 2 ) + 2 cos (
2
) cos (
C−B
2
< 90° ⟹ 2 cos (
C−B
2
)>0
C−B
sin ( 2 ) + 2 cos (
2
)≠0
Yue Kwok Choy
25 June, 2015
3