Trigonometry Mixed Review: Part 1 (3)

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Question 1 of 9

Solve for

$h$

Round your answer to two decimal places

1. $16.92 m$
2. $7.95 m$
3. $6.81 m$
4. $12.17 m$

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Sin Ratio

$sin=\frac{\color{#004ec4}{\text{opposite}}}{\color{#e85e00}{\text{hypotenuse}}}$

Cos Ratio

$cos=\frac{\color{#00880a}{\text{adjacent}}}{\color{#e85e00}{\text{hypotenuse}}}$

Tan Ratio

$tan=\frac{\color{#004ec4}{\text{opposite}}}{\color{#00880a}{\text{adjacent}}}$

First we need to identify which trig ratio to use.

One of the known angles $(29°35′)$ has

$h$ as an $\text(adjacent)$ side and the other length $(14)$ is the $\text(hypotenuse)$

Hence, we can use the

$cos \text(ratio)$ to solve for

$x$

$cos theta$	$=$	$\frac{\color{#00880a}{\text{adjacent}}}{\color{#e85e00}{\text{hypotenuse}}}$	$cos \text(ratio)$

$cos (29°35′)$	$=$	$\frac{\color{#00880a}{h}}{\color{#e85e00}{14}}$	Plug in the values

Now we need to have

$x$ on one side of the equation

$cos (29°35′)$	$=$	$h/14$

$14 times cos (29°35′)$	$=$	$h$	Multiply both sides by $14$
$14 times 0.8696385746$	$=$	$h$	Evaluate $cos (29°35′)$ on the calculator
$12.17$	$=$	$h$	Rounded to two decimal places
$h$	$=$	$12.17$

$h=12.17$

Question 2 of 9

2. Question
Find the area of the Triangle

The given measurements are in centimetres

Round your answer to the nearest whole number
- $\text(Area )=$ (14) $cm^2$
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Area of a Triangle Formula

$\text(Area )=1/2 xx$ $a$ $times$ $c$ $times sin$ $B$

Remember

Uppercase letters represent angles in the triangle

Lowercase letters represent the side lengths

Labelling the triangle

Solve for the area using the Area of a Triangle formula

$\text(Area)$ $=$ $1/2 xx$ $a$ $times$ $c$ $times sin$ $B$ Area of a Triangle formula

$=$ $1/2 xx$ $8$ $times$ $7$ $times sin$ $30°$ Plug in the known lengths

$=$ $14 cm^2$

The given measurements are in centimetres, so the area is measured as square centimetres

$\text(Area)=14 cm^2$

Question 3 of 9

Solve for

$theta$

Round your answer to the nearest degree

$theta=$ (71) $°$

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