# Simple Trigonometry question problem

In the triangle ABC, the height of AH is $11$ cm, and the length of the rib BC is $17$ cm. The angle inbetween the rib AB to height AH is $25^\circ$.

A) Please write an trigonometric expression that expresses the ratio between HB and AH. Done, it is $tan(25)=BH/AH$

B) Calculate the length of BH. Answer: $5.13$ cm.

C) Calculate the size of the angle CAH. <- Im having problems with this one. How ever I found the size of CH, and added BH to it, . but Idk how i would find CA , because A is a X (On the left triangle) thats the issue.

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Angle in between $AB$ and $AH$ is $\*25 cm\*$!! –  Aang Mar 5 '13 at 12:12
I know that, but how about the left triangle? –  Jony Mar 5 '13 at 12:16
@jony an angle isn't measured in cm :) –  Dominic Michaelis Mar 5 '13 at 12:17
Yeah, sorry it's not CM, it says 'o'. –  Jony Mar 5 '13 at 12:21

Knowing the length of AH and CH would be enough to calculate the size of the angle CAH..

And since you already know them, just $tan^{-1}$ the ratio would give you the angle.

$$tan^{-1}(11/11.87) = 42.82...^\circ$$

$tan^{-1}(x)$ is the inverse function of $tan(x)$, also known as arctangent.

$tan^{-1}(x)$ can also be written as $arctan(x)$.

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Thanks about this! but sorry, I haven't really learned that format, what is −1 ? –  Jony Mar 5 '13 at 12:33
@Jony You are welcome. I am not sure exactly why it is written that way (there are definitely a history behind it), but just think of it as a way of writing an inverse function:) The same formate can be applied to $sin$, $cos$ and all other functions. However, bear in mind that $something^{-1}$ would mean different things when it comes to exponentiation. –  Archy Wilhes Mar 5 '13 at 12:48

Since triangle AHC is a right angled triangle, we can calculate angle CAH as,

tan (CAH) = 11.87/11

=> angle(CAH) = $tan^-1 (11.87/11)$

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To find $CA$, simply use Pythagoras' theorem since triangle $ACH$ is right-angled. $$CA^2=CH^2+AH^2$$

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