Please help with both questions thank you~~

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# Question

# Answer

voidage

It’s been a while since I touched on this. Here is my working.

sin θ = O/H = 5/13 (O = Opposite, H = Hypotenuse, A = Adjacent)

Opposite = 5 units

Hypothenuse = 13 units

Adjacent = 12 units (Pythagorean theorem)

1a) tan θ sin θ = (O/A) (O/H) = 5/12 x 5/13 = 25/156

1b) 3 tan θ + cos θ = 3 (5/12) + 12/13 = 5/4 + 12/13 = 65/52 + 48/52 = 113/52 = 2 9/52

1c) cos (90- θ) = sin θ = 5/13

- tan θ = x/y =è O = x, A = y, H = √(x
^{2 }+ y^{2}) - a) cos θ = y/√(x
^{2 }+ y^{2}) - b) sin (90 – θ) = cos θ = y/√(x
^{2 }+ y^{2})

Mazda12345

thank you HAHA i was so dumb

ChiefKiasu

Actually, you are not dumb as the question assumes a right angled triangle

so a^{2} = 13^{2} – 5^{2} = 169-25 =144

=> a=12.

So the triangle’s sides are 5, 12, and 13. You need to keep this in mind.

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