# A few quick math questions

• Nov 27th 2008, 06:06 PM
karate-kid
A few quick math questions
The triangle is a right triangle.
'If AB=7cm,AC=10cm,then what is the measure of angle A?'

'If angle A=38 degrees,AB=10cm,what is the length of BC?'

'If angle A=20 degrees,AC=18cm
i)What is the length of the adjecant side?

ii)What is the length of the opposite side?'

Any help on these would be fantastic
• Nov 27th 2008, 09:57 PM
mr fantastic
Quote:

Originally Posted by karate-kid
The triangle is a right triangle.
'If AB=7cm,AC=10cm,then what is the measure of angle A?'

'If angle A=38 degrees,AB=10cm,what is the length of BC?'

'If angle A=20 degrees,AC=18cm
i)What is the length of the adjecant side?

ii)What is the length of the opposite side?'

Any help on these would be fantastic

Q1 You will need to make it clear at which vertex (A, B or C) the right angle is. Alternatively, make it clear which (if any) of the sides mentioned is the hypotenuse.

Q2 I suppose AC is the hypotenuse. Then $\cos 20 = \frac{\text{adjacent}}{18}$ and $\sin 20 = \frac{\text{opposite}}{18}$.
• Nov 27th 2008, 10:08 PM
karate-kid
For question 1,I think the vertex at which the right angle is is B,and AC is the hypotenuse.
Thanks for the help :)
• Nov 27th 2008, 10:45 PM
mr fantastic
Quote:

Originally Posted by karate-kid
The triangle is a right triangle.
'If AB=7cm,AC=10cm,then what is the measure of angle A?'

'If angle A=38 degrees,AB=10cm,what is the length of BC?'

[snip]

Quote:

Originally Posted by karate-kid
For question 1,I think the vertex at which the right angle is is B,and AC is the hypotenuse.
Thanks for the help :)

$\cos \theta = \frac{7}{10}$. Press some buttons on a calculator to solve for $\theta$.

$\tan 38^{0} = \frac{BC}{10}$ (assuming AB is the adjacent side and BC is the opposite side).
• Nov 27th 2008, 10:54 PM
karate-kid
BC is actually the adjacent side,and AB is the opposite side
• Nov 27th 2008, 10:59 PM
mr fantastic
Quote:

Originally Posted by karate-kid
BC is actually the adjacent side,and AB is the opposite side

You should be able to make the required adjustment to what I posted.
• Nov 27th 2008, 11:04 PM
karate-kid
Yea I'll figure it out,thanks again