Question

[Tiến Nguyễnn]

Answer 1

[HT.Phong (9A5)]

a) Ta có: 

\(sin54^o=\dfrac{y}{3}=>y=3\cdot sin54^o\approx2,4\left(cm\right)\\ =>x=\sqrt{3^2-y^2}=\sqrt{9-2,4^2}\approx1,8\left(cm\right)\)

b) Ta có: 

\(sin32^o=\dfrac{1,5}{y}=>y=\dfrac{1,5}{sin32^o}\approx2,8\left(cm\right)\\ =>x=\sqrt{y^2-1,5^2}=\sqrt{2,8^2-1,5^2}\approx2,4\) 

c) Ta có:

\(tan70^o=\dfrac{y}{0,8}=>y=0,8\cdot tan70^o\approx2,2\left(cm\right)\\ =>x=\sqrt{y^2+0,8^2}=\sqrt{2,2^2+0,8^2}\approx2,3\left(cm\right)\)

Answer 2

[Nguyễn Huy Tú]

Cho tam giác ABC vuông tại A, ^B là góc biết số đo 

a, sinB = y/3 => y  \(\approx\)2,42 cm 

cosB = x/3 => y \(\approx\)1,76 cm 

b, sinB = 1,5/y => y = 1,5/sinB \(\approx\)2,83 cm 

tanB = 1,5/x => x = 1,5/tanB => x \(\approx\)2,4 cm 

c, tanB = y/0,8 => y = 0,8.tanB => y \(\approx\)2,19 cm 

cosB = 0,8/x => x = 0,8/cosB => x \(\approx\)2,34 cm