# Trigonometry: Area of Triangles

• Aug 3rd 2012, 12:49 PM
GrigOrig99
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)
$\frac{1}{2}r^2tan\theta$
• Aug 3rd 2012, 02:12 PM
skeeter
Re: Trigonometry: Area of Triangles
note that $|ad| = r\tan{\theta}$

$A = \frac{1}{2}bh = \frac{1}{2} r \cdot r\tan{\theta}$
• Aug 3rd 2012, 03:21 PM
GrigOrig99
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?
• Aug 3rd 2012, 03:43 PM
Plato
Re: Trigonometry: Area of Triangles
Quote:

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?

$\text{Area}(\Delta CAB)\le \text{Sector-area}(CAB)\le\text{Area}(\Delta CAD)$
• Aug 3rd 2012, 03:55 PM
GrigOrig99
Re: Trigonometry: Area of Triangles
Ok, thank you.