This is a Challenge Question from my Intro to Set Theory Homework. I am to show

Given a triangle $\triangle$ ABC, with side lengths A,B,C and Corresponding angles $\alpha$,$\beta$,$c$. Prove that the following statement is true: $$Area(\triangle ABC)=\frac{1}{2}BC\sin(\alpha) \Rightarrow \sin(\alpha +\beta)=\sin(\alpha)\cos(\beta)+\sin(\beta)\cos(\alpha)$$

The Hint:

Compute the area of $\triangle$ABC is 2 different ways. You may use the usual area formula of a right triangle (it’s a special case of the formula you assume, anyway).

So I determined that "$\Rightarrow$" means that for all $\sin(\alpha +\beta)$, $\frac{1}{2}BC\sin(\alpha)$ must be true. So I can try to show that $\sin(\alpha +\beta)$ is a special case of $\frac{1}{2}BC\sin(\alpha)$.

And because of the hint. I was inclined to set the angle $\alpha$ to 90. Thus, $$\frac{1}{2}BC(1)\Rightarrow(1)\cos(\beta)+\sin(\beta)(0)$$ $$\frac{1}{2}BC\Rightarrow \cos(\beta)$$ $$\frac{1}{2}BC\Rightarrow\frac{C}{A}$$

And I'm left with a statement that doesn't mean anything, and frankly looks wrong. I am also unsure how to use the first part of the hint. The only other method I know for solving area of triangle is Heron's Law, which doesn't seem to fit. But, If I somehow got those two equations to equal, would it even prove the relation? or am I working in the wrong direction.

Any help or pointers are greatly appreciated.

  • 1
    $\begingroup$ I don't understand why it has [set-theory] tag. $\endgroup$ – Hanul Jeon Jan 19 '17 at 8:41
  • $\begingroup$ It's not set theory? The class that assigned this homework is called set theory $\endgroup$ – Jess L Jan 19 '17 at 19:22
  • $\begingroup$ It is a question about Euclidean geometry. It does not related to any form of a set. $\endgroup$ – Hanul Jeon Jan 21 '17 at 7:19

From here:

enter image description here


$(1)$ Let $QN$ be a line that is perpendicular with $PR$.

$(2)$ Then $\text {area}_{\triangle PQR} = \text {area}_{\triangle PQN} +\text {area}_{\triangle RQN}$.

$(3)$ We have: $$\text {area}_{\triangle PQR} = (1/2)rp\sin(A+B)$$ $$\text {area}_{\triangle PQN} = (1/2)rh\sin(A)$$ $$\text {atea}_{\triangle RQN}= (1/2)ph\sin(B)$$

$(4)$ Combing $(2) $ and $(3)$ gives us: $$(1/2)rp\sin(A+B) = (1/2)rh\sin(A) + (1/2)ph\sin(B)$$ $$\Rightarrow rp\sin(A+B) = rh\sin(A) + ph\sin(B)$$ $$ \Rightarrow \sin(A+B) = (\frac {h}{p})\sin(A) + (\frac {h}{r})\sin(B)$$

$(5)$ We also know $\cos B = \frac {h}{p} $ and $\cos A= \frac {h}{r} $.

$(6)$ So that we have: $$\sin(A+B) = \cos(B)\sin(A) +\cos(A)\sin(B)$$

Also $(a) \implies (b)$ means that if $(a) $ is true then $(b) $ must also be true. Hope it helps.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.