Simon's copy of Q4 Solve for x and y on a given triangle and calculate the area

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Simon Thomas 5 years, 10 months ago

Created this as a copy of Q4 Solve for x and y on a given triangle and calculate the area.

Name	Status	Author	Last Modified
Q3 Solve for x and y on a given triangle	draft	TEAME CIT	05/10/2016 14:45
Q4 Solve for x and y on a given triangle and calculate the area	draft	TEAME CIT	08/10/2016 21:39
Q1.1 Pythagoras' Theorem and naming sides of right angled triangle	draft	TEAME CIT	08/10/2016 21:38
Q3: Solve for x and y on a given triangle	draft	Denis Flynn	17/01/2017 13:15
MATH6058 Pythagoras' Theorem and naming sides of right angled triangle	draft	Catherine Palmer	23/10/2017 00:40
MATH6058 Solve for x and y on a right angled triangle	draft	Catherine Palmer	23/10/2017 00:42
MATH6058 Solve for x and y on a given triangle and calculate the area	Ready to use	Catherine Palmer	29/03/2018 11:27
Simon's copy of Q1.1 Pythagoras' Theorem and naming sides of right angled triangle	draft	Simon Thomas	01/05/2019 14:59
Simon's copy of Q1.1 Pythagoras' Theorem	draft	Simon Thomas	01/05/2019 15:44
Simon's copy of Q3 Solve for x and y on a given triangle	draft	Simon Thomas	02/05/2019 09:47
Simon's copy of Q4 Solve for x and y on a given triangle and calculate the area	draft	Simon Thomas	02/05/2019 10:42
Maria's copy of Q1.1 Pythagoras' Theorem and naming sides of right angled triangle	draft	Maria Aneiros	29/05/2019 08:42
Using Pythagoras to find a missing side (Eukleides diagrams)	Ready to use	Deirdre Casey	10/05/2023 09:12
Ann's copy of Using Pythagoras to find a missing side (Eukleides diagrams)	draft	Ann Smith	11/06/2019 10:48

There are 45 other versions that do you not have access to.

Question statement

Consider the sine rule and area formula below. Note that these formulae work for ANY triangle - it does not have to be right-angled.

Note that the sine rule can also be rearranged to give:

$\frac{\sin A}{a} = \frac{\sin B}{b} =\frac{\sin C}{c}$

Now solve the following questions to 2 decimal places:

Note: You may need to scroll down to see the diagrams.

Give any introductory information the student needs.

			Name	Type	Generated Value

ang31a	integer	79
lent31a	integer	80
ans31a	number	64.221
ang32a	integer	52
ans32a	number	61.507
h3a	number	48.468
ans33a	number	1938.720
l2a	number	37.867

			Name	Type	Generated Value

lent11	integer	39
ans11	number	44.984
ans12	number	50.187
lent22	integer	45
lent21	integer	77
ans21	number	23.952
ang31	integer	79
lent31	integer	107
ans31	number	85.895
ang32	integer	49
ans32	number	82.265
h3	number	64.826
ans33	number	3468.191
l2	number	56.352
ang21	integer	136
ang11	integer	73
ang12	integer	59
h1	number	33.430
w1	number	30.101
ans13	number	838.876
ans22	number	20.048
ans23	number	37.999
h2	number	31.260
w2	number	70.369
ans24a	number	1099.867
ans24	number	593.917

Automatically regenerate variables?

Name

Data type

Value

xxxxxxxxxx
 
random(71..85)

JME function reference

Description

Describe what this variable represents, and list any assumptions made about its value.

Can an exam override the value of this variable?

Generated value: `integer`

→ Used by:

ans31a
ans32a

This variable doesn't seem to be used anywhere.

Parts

Part a) Gap-fill
1. Gap Gap 0. Number entry
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2. Gap Gap 1. Number entry
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3. Gap Gap 2. Number entry
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1. Step Information only
Part b) Gap-fill
1. Gap Gap 0. Number entry
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2. Gap Gap 1. Number entry
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3. Gap Gap 2. Number entry
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4. Gap Gap 3. Number entry
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Gap-fill

Ask the student a question, and give any hints about how they should answer this part.

Find the values of $x$ and $y$ , and the area of the triangle:

{tri(ans12,h1,lent11,ang11,ang12,w1)}

$x$ = Gap 0.mm

$y$ = Gap 1.mm

Area = Gap 2. $mm^2$

Advice

(a)

The missing angle is $180 - \var{ang11} - \var{ang12} = \var{180-ang11-ang12}$

Then we can use the sine rule to find $x$ and $y$ :

$\frac{\var{lent11}}{\sin(\var{180-ang11-ang12})} = \frac{x}{\sin(\var{ang12})}$

$x =\sin(\var{ang12}) \times \frac{\var{lent11}}{\sin(\var{180-ang11-ang12})} = \var{ans11}$

$\frac{\var{lent11}}{\sin(\var{180-ang11-ang12})} = \frac{y}{\sin(\var{ang11})}$

$y =\sin(\var{ang11}) \times \frac{\var{lent11}}{\sin(\var{180-ang11-ang12})} = \var{ans12}$

To find the area we use the formula $\text{Area} = \frac{1}{2}ab\sin C$

$\text{Area} = \frac{1}{2} \times \var{ans11} \times \var{lent11} \times \sin(\var{ang11}^{\circ}) = \var{ans13}$ mm $^2$

(b)

First use the rearranged sine rule to find $X$ :

$\frac{\sin(\var{ang21})}{\var{lent21}} = \frac{\sin(X)}{\var{lent22}}$

$\sin(X) =\var{lent22} \times \frac{\sin(\var{ang21})}{\var{lent21}}$

$X =\sin ^{-1} (\var{lent22} \times \frac{\sin(\var{ang21})}{\var{lent21}}) = \var{ans21}^{\circ}$

Then we can find $Y$ by considering angles in a triangle:

$Y = 180-\var{ans21}-\var{ang21} = \var{ans22}^{\circ}$

Finally use the sine rule again to find $z$ :

$\frac{\var{lent21}}{\sin(\var{ang21})} = \frac{z}{\sin(\var{ans22})}$

$z = \frac{\var{lent21}}{\sin(\var{ang21})} \times \sin(\var{ans22}) = \var{ans23}$ mm

To find the area we use the formula $\text{Area} = \frac{1}{2}ab\sin C$ :

$\text{Area} =\frac{1}{2} \times \var{ans23} \times \var{lent22} \times \sin(\var{ang21}) = \var{ans24}$ mm $^2$

Give a worked solution to the whole question.

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This question is used in the following exam:

Trigonometry and Pythagoras by Simon Thomas in Maths support.

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Simon's copy of Q4 Solve for x and y on a given triangle and calculate the area

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Taxonomy: mathcentre

Taxonomy: Kind of activity

Taxonomy: Context

Contributors

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Simon Thomas 5 years, 10 months ago

Error in variable testing condition

Error

Warning

Generated value: `integer`

Parts

Gap-fill

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Error

Simon's copy of Q4 Solve for x and y on a given triangle and calculate the area

Metadata

Taxonomy: mathcentre

Taxonomy: Kind of activity

Taxonomy: Context

Contributors

History

Comment

Checkpoint description

Simon Thomas 5 years, 10 months ago

Error in variable testing condition

Error

Warning

Generated value: integer

Parts

Gap-fill

Generated value: `integer`