Properties AND Solution OF Triangles flash cards

1. 1.State the Law of Sines for a triangle with sides $a,b,c$ opposite angles $A,B,C$ and circumradius $R$.

   $\dfrac{a}{\sin A}=\dfrac{b}{\sin B}=\dfrac{c}{\sin C}=2R$

   Hint: Involves circumradius $R$

2. 2.Write the projection formula for side $a$ in terms of $b, c$ and the angles.

   $a=b\cos C+c\cos B$

   Hint: Sum of projections of other two sides

3. 3.What is the formula for the area of a triangle using $R$, and using the exradius/inradius relation $\Delta = rs$?

   $\Delta=\dfrac{abc}{4R}=rs=r_1(s-a)=r_2(s-b)=r_3(s-c)$, where $s=\dfrac{a+b+c}{2}$

   Hint: Also $\Delta=\sqrt{s(s-a)(s-b)(s-c)}$ (Heron's)

4. 4.State Napier's Analogy (tangent rule) relating $\dfrac{B-C}{2}$ to the sides and angle $A$.

   $\tan\left(\dfrac{B-C}{2}\right)=\dfrac{b-c}{b+c}\cot\left(\dfrac{A}{2}\right)$

   Hint: Half-angle difference formula

5. 5.In a triangle, express the m-n theorem: if point $D$ divides $BC$ such that $BD:DC=m:n$ and $\angle ADB=\theta$, $\angle ADC=\phi$ (with $\angle BAD=\alpha$, $\angle DAC=\beta$), state the relation.

   $(m+n)\cot\theta = m\cot\alpha - n\cot\beta$ and $(m+n)\cot\theta = n\cot B - m\cot C$

   Hint: Used for cevian angle problems

6. 6.In $\triangle ABC$, $a=6$, $b=8$ and $\angle A=30°$. How many distinct triangles satisfy these conditions, and what is $\sin B$?

   By the sine rule, $\sin B=\dfrac{b\sin A}{a}=\dfrac{8\times \frac12}{6}=\dfrac{2}{3}$. Since $b\sin A=4two triangles exist, with$B\approx41.8°$or$B\approx138.2°$(supplementary values of$\sin^{-1}\frac23$).

   Hint: Compare $a$ with $b\sin A$ and $b$ to decide how many solutions SSA gives.

7. 7.For triangle $ABC$ with sides $a,b$ and an acute angle $A$ given (SSA), state the precise condition on $a$ (relative to $b\sin A$ and $b$) for exactly two triangles to exist.

   Exactly two triangles exist iff $b\sin A

   Hint: Think about how many times a circle of radius $a$ centred at $C$ meets the ray from $B$.

8. 8.In $\triangle ABC$, if $\dfrac{\cos A}{a}=\dfrac{\cos B}{b}=\dfrac{\cos C}{c}$, identify the triangle's type.

   Using $a=2R\sin A,\,b=2R\sin B,\,c=2R\sin C$, the condition becomes $\dfrac{\cos A}{\sin A}=\dfrac{\cos B}{\sin B}=\dfrac{\cos C}{\sin C}$, i.e. $\cot A=\cot B=\cot C$. Hence $A=B=C$, so the triangle is equilateral.

   Hint: Replace $a,b,c$ using the sine rule to turn this into a statement about cotangents.

9. 9.Prove that in any $\triangle ABC$: $a\sin(B-C)+b\sin(C-A)+c\sin(A-B)=0$.

   Write $a=2R\sin A$ etc. The sum becomes $2R\big[\sin A\sin(B-C)+\sin B\sin(C-A)+\sin C\sin(A-B)\big]$. Using $A=\pi-(B+C)$ so $\sin A=\sin(B+C)$, each product-pair telescopes via sum-to-product identities and the total cancels to $0$ — confirmed directly, e.g. for $A=90°,B=60°,C=30°$ the three terms are $0.5,-0.75,0.25$ which sum to $0$.

   Hint: Substitute $a,b,c$ via the sine rule and use $A+B+C=\pi$.

10. 10.In $\triangle ABC$, $\angle A=45°$, $\angle B=60°$ and $a=\sqrt2$. Find side $c$ in simplest surd form.

    $C=180°-45°-60°=75°$, and $\sin75°=\dfrac{\sqrt6+\sqrt2}{4}$. By the sine rule, $\dfrac{a}{\sin A}=\dfrac{\sqrt2}{\sin45°}=2=2R$. So $c=2R\sin C=2\cdot\dfrac{\sqrt6+\sqrt2}{4}=\dfrac{\sqrt6+\sqrt2}{2}$.

    Hint: Find the third angle first, then use $2R=a/\sin A$.

Open the interactive deck for the other 75 cards, with self-grading so the ones you keep missing come back.