As with any triangle, the sum of angles is 180°.
(a)
(x -5) +(3x +30) +35 = 180 . . . . . sum of angles is 180
4x +60 = 180 . . . . . . . . . . . . . . . . collect terms
4x = 120 . . . . . . . . . . . . . . . . . . . . subtract 60
x = 30 . . . . . . . . . . . . . . . . . . . . . . divide by 4
(b)
Angle A = x -5 = 30 -5 = 25
Angle B = 3x +30 = 3*30 +30 = 120
Answer:
I think it is the right clock wise?
Some basic formulas involving triangles
\ a^2 = b^2 + c^2 - 2bc \textrm{ cos } \alphaa 2 =b 2+2 + c 2
−2bc cos α
\ b^2 = a^2 + c^2 - 2ac \textrm{ cos } \betab 2=
m_b^2 = \frac{1}{4}( 2a^2 + 2c^2 - b^2 )m b2 = 41(2a 2 + 2c 2-b 2)
b
Bisector formulas
\ \frac{a}{b} = \frac{m}{n} ba =nm
\ l^2 = ab - mnl 2=ab-mm
A = \frac{1}{2}a\cdot b = \frac{1}{2}c\cdot hA=
\ A = \sqrt{p(p - a)(p - b)(p - c)}A=
p(p−a)(p−b)(p−c)
\iits whatever A = prA=pr with r we denote the radius of the triangle inscribed circle
\ A = \frac{abc}{4R}A=
4R
abc
- R is the radius of the prescribed circle
\ A = \sqrt{p(p - a)(p - b)(p - c)}A=
p(p−a)(p−b)(p−c)
378 because 49*3 and then add 35
