A geometry problem by Mark Clancy
Geometry
Level
2
a, b are angles in the first quadrant such that tan(a)=3/4 and tan(b) = 5/12
Find sin(a+b)
56/65
33/65
63/65
8/9
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1 solution
Sin(a+b)= sin(a) cos(b) + cos(a) sin(b)
1) Consider a right angled triangle with side ratios- 3, 4 and 5 (where 5 is the hypotenuse and is found by pythogorus theorem) and an angle, a,
Where tan(a)= 3/4,
Therefore: sin(a)= 3/5 and cos(a)= 4/5
2) Consider a right angled triangle with side ratios- 5, 12 and 13 (where 13 is the hypotenuse and is found by pythogorus theorem) and an angle, b,
Where tan(b) = 5/12,
Therefore: sin(b)= 5/13 and cos(b)= 12/13
Sin(a+b)= (3* 12)/ (5* 13) + (4* 5)/ (5* 13)= (36+20)/65
= 56/65
OR
sin[arctan(3/4) + arctan(5/12)]