All courses Math III · G-SRT.9 44 of 55
Derive the triangle area formula A=1/2ab sin(C) using an auxiliary altitude.

Find the area of a triangle from two sides and the included angle

Problem
Which area is correct for a triangle with sides 8 and 10 and included angle 30 degrees?
Triangle with side lengths 8 and 10 meeting at a 30-degree included angle and a dashed perpendicular altitude labeled h. Open full size
Your answer
Choose an answer
With a free account

See progress instead of guessing.

Gozunta keeps your results together so you can see what you have practiced and how it went.

Create a free account
With paid access

Follow a skill back to its prerequisites.

All-course access lets you move between Math Foundations and Math I–III when an earlier idea needs attention.

Compare plans

Hint

Use the triangle area formula with the two given sides and their included angle: \(A=\tfrac12 ab\sin(C)\).

For a triangle with sides 8 and 10 and included angle \(30^\circ\), use \(A=\tfrac12(8)(10)\sin(30^\circ)\). Also, \(\sin(30^\circ)=\tfrac12\).

Solution walkthrough

01

Identify the correct area formula

\[A~=~\tfrac12~\text{ab}\sin(C)\]

You are given two sides, 8 and 10, and the included angle, \(30^\circ\). That is exactly when the trigonometric area formula applies.

02

Substitute the given values

\[A~=~\tfrac12(8)(10)\sin(30^\circ)\]

Here, \(a=8\), \(b=10\), and the included angle is \(C=30^\circ\). Plug those directly into the formula.

03

Evaluate the sine and simplify

\[A~=~\tfrac12(8)(10)~\left(\tfrac12\right)~=~\tfrac12(80)~\left(\tfrac12\right)~=~20\]

Since \(\sin(30^\circ)=\tfrac12\), the multiplication becomes straightforward. First, \(8\cdot 10=80\), and then \(\tfrac12\cdot 80\cdot \tfrac12=20\).

04

State the area

\[\text{area }~=~20\]

So, using the trigonometric area formula for sides 8 and 10 with included angle \(30^\circ\), the area is 20.

+

Another way

  1. You can verify by finding an altitude first: if 10 is the base, then the height is \(8\sin(30^\circ)=4\). Then \(A=\tfrac12(10)(4)=20\).

  2. You could also use 8 as the base. Then the height is \(10\sin(30^\circ)=5\), so \(A=\tfrac12(8)(5)=20\).

!

Common mistake

A common mistake is to ignore the sine and do \(\tfrac12(8)(10)=40\). The formula needs the included angle factor, so you must multiply by \(\sin(30^\circ)=\tfrac12\), which gives 20 instead.

The same triangle with height 4 and the resulting area 20 shown only in answer support.