# how to find the other angles of an oblique triangle, given two sides and one angle [closed]

If I know the length of two sides of an acute or obtuse triangle and know the angle between these two sides, how can I get one of the missing two angles?

(Obviously, the final missing angle = 180 degrees - the known angle - one of the missing angles.)

Triangle is points $A, B, C$.

Known sides: $\overline{AB}$ & $\overline{BC}$.

Known angle: $\angle ABC$.

## closed as off-topic by Daniel, daw, kingW3, user147263, Tim RaczkowskiSep 26 '15 at 23:36

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Daniel, daw, Tim Raczkowski
If this question can be reworded to fit the rules in the help center, please edit the question.

• Are we told the relationship of the given angle to the two given sides? Such as the sides are $17$ and $19$ and the angle between them is $40^\circ$. Or the angle opposite the side of length $17$ is $40^\circ$? Or are we given no relationship information? – André Nicolas Sep 26 '15 at 13:03
• Angle is between the two sides. – Doug Null Sep 29 '15 at 16:33
• Call the sides $a$, $b$, and $c$. Let the angle opposite side $a$ be $A$, and so on. We are given say sides $a$, $b$ and $\angle C$ between them. Use the Cosine Law $c^2=a^2+b^2-2ab \cos C$ to find $c$. Now we know all the sides, so we can use $a^2=b^2+c^2-2bc\cos A$ to find $A$. Alternately, once we know $c$ choose the smaller of the two sides $a$ and $b$. Say it is $a$. Find $\sin A$ from $\frac{\sin A}{a}=\frac{\sin C}{c}$. Once we know $\sin A$, we know $A$ since $A$ must be acute. – André Nicolas Sep 29 '15 at 17:02

## 3 Answers

Hint: Cosine rule:

Let us say that the sides of a triangle have a length $a,b,c$. And $\alpha$ is the opposite angle of the $A$, then: $$a^2=b^2+c^2-2bc\cos(\alpha).$$

If the angle in between sides is known use first cosine rule, then sine rule.

If the angle in between sides is not known, but opposite to one of them only, then we have $two$ solutions because of the ambiguity

Draw a ray with known side and known angle. Cut the ray twice with known side as radius of circle.

If the angle you know is facing one of the sides you know, then it's best to use the Sine Rule. If the angle is between the two sides you know, then you have to use the Cosine Rule first to find the third side. From there you can continue with either the Sine or Cosine Rule.

Let's say you have a triangle with sides $a$, $b$, and $c$ and angles $A$, $B$, and $C$ (Side $a$ faces angle $A$, side $b$ faces angle $B$ and side $c$ faces angle $C$)...

The Law of Sines (also called Sine Rule) states that: $$\frac{a}{\sin(A)}=\frac{b}{\sin(B)}=\frac{c}{\sin(C)}$$

If you substitute the values that you know, you can find the missing angles by cross multiplication.

The Law of Cosines (also called Cosine Rule) states that: $$c^2 = a^2 + b^2 − 2ab\cos(C)$$ ; where angle $C$ is facing side $c$ and is between the two sides $a$ and $b$.

Here if you substitute the values you know, you can find the third side $c$ facing the angle $C$. And then you can use the Sine or Cosine rule to find the missing angles.

NOTE that these examples are general as I do not know the details of your question. So try and understand the rules, and from there decide which you can use to find your answer.