# Trigonometry problem

• Jun 4th 2007, 04:08 PM
Trentt
Trigonometry problem
Not sure how to do this problem:

Find the other side and angle measures for the given right triangle:
Name: HPJ
Right Angle: H
Given Data: m<J =29°51', j = 4651

Thanks a lot.
• Jun 4th 2007, 05:39 PM
Soroban
Hello,Trentt!

Quote:

Find the other sides and angles.
Triangle $PHJ$
Right Angle: $H$
Given data: . $\angle J = 29^o51',\;j = 4651$

Code:

      P       *       |  *       |    *  h  4651 |        *       |          *       |      29.85°-*       * - - - - - - - - *     H        p          J
We know that: . $\angle H = 90^o$

We have: . $\angle J \:=\:29^o51' \:=\:29 +\frac{51}{60} \:=\:29.85^o$

Then: . $\angle P \:=\:90^o - 29.85^o \:=\:60.15^o$

We have: . $\begin{array}{ccccccc}J & = & 29.85^o & \qquad & k & = & 4651 \\
P & = & \boxed{60.15^o} & & p & = & \boxed{\qquad} \\
H & = & \boxed{90.00^o} & & h & = & \boxed{\qquad} \end{array}$

$\tan P \:= \:\frac{p}{j}\quad\Rightarrow\quad p \:= \:j\!\cdot\tan P \:=\:4651\!\cdot\!\tan60.15^o \:\approx\:8104.7$

$\sin J \:=\:\frac{j}{h}\quad\Rightarrow\quad h \:=\:\frac{j}{\sin J} \:=\:\frac{4651}{\sin29.85^o} \:\approx\:9344.4$

Therefore: . $\begin{array}{ccccccc}J & = & 29.85^o & \qquad & k & = & 4651.0 \\
P & = & \boxed{60.15^o} & & p & = & \boxed{8104.7} \\
H & = & \boxed{90.00^o} & & h & = & \boxed{9344.4} \end{array}$