# Thread: Trigonometry: Area of Triangles

1. ## Trigonometry: Area of Triangles

Based on the books answers, (i) & (ii) are correct but I'm confused about (ii), seeing as how the answer is provided in the question!! Am I meant to prove the statement through the information provided...?
The book's answer to (iii) is provided below, but the confusion for me here is the application of tan, in defining the area of the triangle. The 1st attempt in (iii) is what I expected the answer to be but, based on the books final answer, I made a second attempt to try and account for tan. Also, again based on the books answer, I'm assuming |ad| is meant to = r but I'm unsure as to how I prove this. Can anyone help me out here?

Many thanks.

Q.
A triangle is inscribed in a sector of a circle, center c, radius r, $\theta<90^o$. A right-angled triangle cad circumscribes the sector, as shown. If the area of a sector is $\frac{1}{2}r^2\theta$, find the area of (i) $\triangle cab$, (ii) the sector cab, (iii) $\triangle$ cad. Hence show that $sin\theta<\theta.

Attempt: (i) Area of $\triangle abc$: $\frac{1}{2}(a)(b)(sinC)$
|ac| = r = |bc|, where both lines are connected from the center to the circumference of the circle whose sector is cab
Thus, $\frac{1}{2}(a)(b)(sinC)$ = $\frac{1}{2}r^2sin\theta$

(ii) $\frac{1}{2}r^2\theta$

(iii) Area of $\triangle cad$: $\frac{1}{2}(d)(a)(sinA)$ = $\frac{1}{2}(r)(|cd|)(sin90^o)$ = $\frac{1}{2}(r)(a)(1)$ = $\frac{1}{2}ra$
or...?
$\frac{1}{2}(c)(d)(tan\theta)$ = $\frac{1}{2}(|ab|)(r)(tan\theta)$

Ans.: (From text book): (iii)
$\frac{1}{2}r^2tan\theta$

2. ## Re: Trigonometry: Area of Triangles

note that $|ad| = r\tan{\theta}$

$A = \frac{1}{2}bh = \frac{1}{2} r \cdot r\tan{\theta}$

3. ## Re: Trigonometry: Area of Triangles

Thanks.

Ok, so if I'm to demonstrate $sin\theta<\theta, and knowing that $\theta<90^o$, I can simply pick any degree from $0^0-89^0$ and display the results e.g. $\theta=60^0$ in which case $sin\60^o<\60^o = $0.86<60<1.73$. Although, this leaves me with the problem of the 60 in the middle.

Is there a better way to prove the statement?

4. ## Re: Trigonometry: Area of Triangles

Originally Posted by GrigOrig99
Thanks.

Ok, so if I'm to demonstrate $sin\theta<\theta, and knowing that $\theta<90^o$, I can simply pick any degree from $0^0-89^0$ and display the results e.g. $\theta=60^0$ in which case $sin\60^o<\60^o = $0.86<60<1.73$. Although, this leaves me with the problem of the 60 in the middle.
Is there a better way to prove the statement?
$\text{Area}(\Delta CAB)\le \text{Sector-area}(CAB)\le\text{Area}(\Delta CAD)$

5. ## Re: Trigonometry: Area of Triangles

Ok, thank you.