Question

cp pre - calculus
4.3 (right triangle trig) quiz review
name
#1 - 6 state the six trig functions in the simplest form. no decimals. no radicals in denominators.
#7 - 11. state the other 5 trig functions given that \\(\cot\theta = 9\\). no decimals. no radicals in denominators.
#12 - 17. solve the following triangles. round degrees to the tenth and sides to the hundredth.
(pictures not drawn to scale.)
angles:
a = 39.2
b = 50.8
c =
sides:
a =
b = 2.1
c =
angles:
a =
b =
c = 90
sides:
a = 2
b = 7
c =

Answer:

For #1 - 6 (Right Triangle with Opposite = 5, Hypotenuse = 7, Adjacent = $\sqrt{24}$)

Step 1: Recall Trigonometric Ratios

For an acute angle $\theta$ in a right triangle:

• $\sin\theta=\frac{\text{opposite}}{\text{hypotenuse}}$
• $\cos\theta=\frac{\text{adjacent}}{\text{hypotenuse}}$
• $\tan\theta=\frac{\text{opposite}}{\text{adjacent}}$
• $\csc\theta=\frac{\text{hypotenuse}}{\text{opposite}}$
• $\sec\theta=\frac{\text{hypotenuse}}{\text{adjacent}}$
• $\cot\theta=\frac{\text{adjacent}}{\text{opposite}}$

Given: Opposite = 5, Adjacent = $\sqrt{24}$, Hypotenuse = 7.

Step 2: Calculate $\sin\theta$

$\sin\theta=\frac{\text{opposite}}{\text{hypotenuse}}=\frac{5}{7}$

Step 3: Calculate $\cos\theta$

$\cos\theta=\frac{\text{adjacent}}{\text{hypotenuse}}=\frac{\sqrt{24}}{7}=\frac{2\sqrt{6}}{7}$ (Simplify $\sqrt{24}=\sqrt{4\times6}=2\sqrt{6}$)

Step 4: Calculate $\tan\theta$

$\tan\theta=\frac{\text{opposite}}{\text{adjacent}}=\frac{5}{\sqrt{24}}=\frac{5\sqrt{24}}{24}=\frac{5\times2\sqrt{6}}{24}=\frac{5\sqrt{6}}{12}$ (Rationalize the denominator: multiply numerator and denominator by $\sqrt{24}$)

Step 5: Calculate $\csc\theta$

$\csc\theta=\frac{\text{hypotenuse}}{\text{opposite}}=\frac{7}{5}$

Step 6: Calculate $\sec\theta$

$\sec\theta=\frac{\text{hypotenuse}}{\text{adjacent}}=\frac{7}{\sqrt{24}}=\frac{7\sqrt{24}}{24}=\frac{7\times2\sqrt{6}}{24}=\frac{7\sqrt{6}}{12}$ (Rationalize the denominator)

Step 7: Calculate $\cot\theta$

$\cot\theta=\frac{\text{adjacent}}{\text{opposite}}=\frac{\sqrt{24}}{5}=\frac{2\sqrt{6}}{5}$ (Simplify $\sqrt{24}$)

For #7 - 11 (Given $\cot\theta = 9$)

Recall that $\cot\theta=\frac{\text{adjacent}}{\text{opposite}}=\frac{9}{1}$ (We can consider a right triangle where adjacent = 9, opposite = 1, then hypotenuse $=\sqrt{9^2 + 1^2}=\sqrt{82}$)

Step 1: Find $\tan\theta$

$\tan\theta=\frac{1}{\cot\theta}=\frac{1}{9}$

Step 2: Find $\sin\theta$

$\sin\theta=\frac{\text{opposite}}{\text{hypotenuse}}=\frac{1}{\sqrt{82}}=\frac{\sqrt{82}}{82}$ (Rationalize the denominator)

Step 3: Find $\cos\theta$

$\cos\theta=\frac{\text{adjacent}}{\text{hypotenuse}}=\frac{9}{\sqrt{82}}=\frac{9\sqrt{82}}{82}$ (Rationalize the denominator)

Step 4: Find $\csc\theta$

$\csc\theta=\frac{\text{hypotenuse}}{\text{opposite}}=\sqrt{82}$

Step 5: Find $\sec\theta$

$\sec\theta=\frac{\text{hypotenuse}}{\text{adjacent}}=\frac{\sqrt{82}}{9}$

For #12 - 17 (First Triangle: $A = 39.2^\circ$, $B = 50.8^\circ$, $C = 90^\circ$, $b = 2.1$)

Step 1: Find side $a$ (opposite $A$)

We know that $\tan A=\frac{a}{b}$ (since $\tan\theta=\frac{\text{opposite}}{\text{adjacent}}$)
So, $a = b\times\tan A$
$\tan(39.2^\circ)\approx0.81$ (Using calculator)
$a = 2.1\times0.81\approx1.70$

Step 2: Find side $c$ (hypotenuse)

We can use Pythagoras theorem: $c=\sqrt{a^2 + b^2}$
$a\approx1.70$, $b = 2.1$
$a^2=(1.70)^2 = 2.89$, $b^2=(2.1)^2 = 4.41$
$c=\sqrt{2.89 + 4.41}=\sqrt{7.3}\approx2.70$
Or use $\sin A=\frac{a}{c}\implies c=\frac{a}{\sin A}$
$\sin(39.2^\circ)\approx0.63$
$c=\frac{1.70}{0.63}\approx2.70$

For #12 - 17 (Second Triangle: $C = 90^\circ$, $a = 2$, $b = 7$)

Step 1: Find side $c$ (hypotenuse)

Using Pythagoras theorem: $c=\sqrt{a^2 + b^2}=\sqrt{2^2 + 7^2}=\sqrt{4 + 49}=\sqrt{53}\approx7.28$

Step 2: Find angle $A$

$\tan A=\frac{a}{b}=\frac{2}{7}\approx0.2857$
$A=\arctan(0.2857)\approx15.9^\circ$

Step 3: Find angle $B$

Since $A + B + C = 180^\circ$ and $C = 90^\circ$
$B = 90^\circ - A\approx90^\circ - 15.9^\circ = 74.1^\circ$

Final Answers (Summarized)

#1 - 6:

1. $\sin\theta=\boldsymbol{\frac{5}{7}}$
2. $\csc\theta=\boldsymbol{\frac{7}{5}}$
3. $\cos\theta=\boldsymbol{\frac{2\sqrt{6}}{7}}$
4. $\sec\theta=\boldsymbol{\frac{7\sqrt{6}}{12}}$
5. $\tan\theta=\boldsymbol{\frac{5\sqrt{6}}{12}}$
6. $\cot\theta=\boldsymbol{\frac{2\sqrt{6}}{5}}$

#7 - 11:

1. $\tan\theta=\boldsymbol{\frac{1}{9}}$
2. $\sin\theta=\boldsymbol{\frac{\sqrt{82}}{82}}$
3. $\cos\theta=\boldsymbol{\frac{9\sqrt{82}}{82}}$
4. $\csc\theta=\boldsymbol{\sqrt{82}}$
5. $\sec\theta=\boldsymbol{\frac{\sqrt{82}}{9}}$

#12 - 17 (First Triangle):

• $a\approx\boldsymbol{1.70}$
• $c\approx\boldsymbol{2.70}$

#12 - 17 (Second Triangle):

• $A\approx\boldsymbol{15.9^\circ}$
• $B\approx\boldsymbol{74.1^\circ}$
• $c\approx\boldsymbol{7.28}$