Jawab pertanyaan berikut

AB = c = 13
AC = b = 13
BC = a = 10
s = 1/2 (a + b + c) = 1/2 (10 + 13 + 13) = 1/2 (36) = 18
Luas ∆ = √(s (s - a) (s - b) (s - c))
= √(18 (18 - 13) (18 - 13) (18 - 10))
= √(18 . 5 . 5 . 8)
= √3600
= 60

4) r = L/s = 60/18 = 10/3 => r = OP = OR = OQ = 10/3

1) AP = √(AC^2 - PC^2) = √(13^2 - 5^2) = √(169 - 25) = √144 = 12
AO = AP - OP = 12 - 10/3 = 26/3
AR = √(AO^2 - OR^2)
= √((26/3)^2 - (10/3)^2)
= √(676/9 - 100/9)
= √(576/9)
= 24/3
= 8

Atau AR = s - a = 18 - 10 = 8

2) sin
3) Tan
5) cos AB . AO = |AB| . |AO| cos
Sudut ABC = sudut ABP = x
Sudut antara AB dan BC = 180° - x
cos x = BP/AB = 5/13
AB . BC = |AB| |BC| cos (180° - x)
= 13 . 10 . - cos x
= 130 . -5/13
= -50

[tex]\displaystyle \text{AP}^2=\text{AC}^2-\text{CP}^2\\\text{AP}^2=13^2-5^2\\\text{AP}=12\\\\\text{OP}=\frac{L}{s}\\\text{OP}=\frac{L}{\frac12K}\\\text{OP}=\frac{60}{\frac12(36)}\\\text{OP}=\frac{60}{18}\\\boxed{\boxed{\text{OP}=\text{OR}=\text{OQ}=\frac{10}{3}}}\\\\\text{AR}^2=\text{AO}^2-\text{OR}^2\\\text{AR}^2=(\text{AP}-\text{OP})^2-\text{OR}^2\\\text{AR}^2=\text{AP}^2-2\text{AP}\cdot\text{OP}+\text{OP}^2-\text{OR}^2\\\text{AR}^2=12^2-2\cdot12\cdot\frac{10}{3}\\\text{AR}^2=64\\\boxed{\boxed{\text{AR}=8}}[/tex]

[tex]\displaystyle \sin\angle\text{AOR}=\frac{\text{AR}}{\text{AO}}\\\sin\angle\text{AOR}=\frac{8}{12-\frac{10}{3}}\\\sin\angle\text{AOR}=\frac{8}{\frac{36-10}{3}}\\\sin\angle\text{AOR}=\frac{8}{\frac{26}{3}}\\\boxed{\boxed{\sin\angle\text{AOR}=\frac{12}{13}}}\\\\\tan\angle\text{AOR}=\frac{\text{AR}}{\text{OR}}\\\tan\angle\text{AOR}=\frac{8}{\frac{10}{3}}\\\boxed{\boxed{\tan\angle\text{AOR}=\frac{12}{5}}}[/tex]

[tex]\displaystyle \vec{\text{AB}}\cdot\vec{\text{AO}}=AB\cdot\text{AO}\cos\angle\text{BAO}\\\vec{\text{AB}}\cdot\vec{\text{AO}}=13\cdot\frac{26}{3}\cdot\frac{12}{13}\\\vec{\text{AB}}\cdot\vec{\text{AO}}=26\cdot4\\\boxed{\boxed{\vec{\text{AB}}\cdot\vec{\text{AO}}=104}}[/tex]
[tex]\displaystyle \vec{\text{AB}}\cdot\vec{\text{BC}}=AB\cdot\text{BC}\cos\angle\text{ABC}\\\vec{\text{AB}}\cdot\vec{\text{BC}}=13\cdot10\cdot\frac{5}{13}\\\vec{\text{AB}}\cdot\vec{\text{BC}}=10\cdot5\\\boxed{\boxed{\vec{\text{AB}}\cdot\vec{\text{BC}}=50}}[/tex]