Tutor profile: Akash K.

In a right triangle ABC, tan(A) = 3/4. Find sin(A) and cos(A).

According to pythagoras theorem, h^2=b^2+p^2 Since tan(A)=p/b here, p=3 and b=4 Putting these values in above pythagoras therorem, h^2=(4)^2+(3)^2 h^2=16+9 h^2=25 taking square root on both sides, h=5 Now, since sin(A)=p/h So, sin(A)=3/5 and cos(A)=b/h so, cos(A)=4/5

What is the equation of a vertical line passing through point (-4,7)?

Since equation of a vertical line is in the form of x=a, So in the given co=ordinates, since a=-4 and b=7 The equation will be x=-4

-10x - 19 = 19 - 8x

-10x - 19 =19 - 8x Step -1) First we will bring all x's on Left hand side and all constants(without any vaiable) on Right hand side -10x-(-8x)=19-(-19) Step-2) since -,- adds up to +,-8x will become 8x -10x+8x=19+19 Step -3) Now apply simple BODMAS rule to solve both LHS and RHS -2x=38 Step -4) Divide both sides by -2 -2x/(-2) =38/(-2) So the answer will be: x=- -19