Question

2
If in the triangle shown above all three interior
angles are multiples of 10 and angle ACB is neither
the largest nor the smallest angle, how many distinct
values of x are possible?
o
B 1
© 2
D3
© More than 3​

Answer 1

Answer: (B) 1

Concept:  In  △ABC

∠ABC +∠CAB +∠ACB = 180°

Given: In  △ABC all angles are multiple of 10.

∠ACD = 3x

∠CAB = 2x

∠ACB is neither smallest nor largest angle of triangle.

To find: Possible values of x.

Step-by-step explanation:

△ABC all angles are multiple of 10.

∠ACD = 3x

∠CAB = 2x

∠ACB is neither smallest nor largest angle of triangle.

In  △ABC

∠ABC +∠CAB +∠ACB = 180°     ---------(1)

∠ACD and ∠ACB are supplementary angles

∠ACD + ∠ACB = 180°   ---------------------(2)

From equation 1 and 2

∠ABC +∠CAB +∠ACB  = ∠ACD + ∠ACB

∠ABC + 2x = 3x

∠ABC = x

In  △ABC

∠CAB = 2x

∠ABC = x

∠ACB = 180° - 3x

Now angles are multiple of 10

Hence x should also be multiple of 10

Putting x = 10

∠CAB = 20

∠ABC = 10

∠ACB = 150

angle ACB is largest

x is not equal to 10

Putting x = 20

∠CAB = 40

∠ABC = 20

∠ACB = 80

angle ACB is largest

x is not equal to 20

Putting x = 30

∠CAB = 60

∠ABC = 30

∠ACB = 90

angle ACB is largest

x is not equal to 30

Putting x = 40

∠CAB = 80

∠ABC = 40

∠ACB = 60

angle ACB is neither largest nor smallest

x = 40

Putting x = 50

∠CAB = 100

∠ABC = 50

∠ACB = 30

angle ACB is smallest

x is not equal to 50

Hence we got only one value of x for which all conditions are satisfied

Answer = 1

#SPJ1

Answer 2

Answer:

The answer is 1

Any triangle's three angles always sum to 180°. Therefore, if you only have two angles available, add them together, and then deduct the total from 180°.

step-by-step explanation:

△A, B, and C are all multiples of 10.

∠$ACD=3x$

∠$CAB=2x$

Neither the lowest nor the biggest angle in a triangle is ∠$ACB$.

A triangle's angles add up to $180$°.

In  △$ABC$

∠$ABC+$∠$CAB+$∠$ACB=180$° $------(i)$

The supplementary angles ∠$ACD$ and ∠$ACB$

∠$ACD+$ ∠$ACB=180$° $------(ii)$

based on equations $(i)and(ii)$

∠$ABC+$∠$CAB+$∠$ACB=$ ∠$ACD+$ ∠$ACB$

∠$ABC+$ $2x=3x$

∠$ABC=x$

In  △$ABC$

∠$CAB=2x$

∠$ABC=x$

∠$ACB=180$° $-3x$

Angles are now multiples of $10.$

Consequently, $x$ must likewise be a multiple of $10.$

adding $10$ to $x$

∠$CAB=20$

∠$ABC=10$

∠$ACB=150$

∠$ACB$ is the biggest

$x$ doesn't equal $10$

adding $20$ to $x$

∠$CAB=40$

∠$ABC=20$

∠$ACB=80$

∠$ACB$ is the biggest

$x$ doesn't equal $20$

adding $30$ to $x$

∠$CAB=60$

∠$ABC=30$

∠$ACB=90$

∠$ACB$ is the biggest

$x$ doesn't equal $30$

adding $40$ to $x$

∠$CAB=80$

∠$ABC=40$

∠$ACB=60$

neither the largest nor the smallest angle is ∠$ACB$

$x=40$

adding $50$ to $x$

∠$CAB=100$

∠$ABC=50$

∠$ACB=30$

the smallest ∠$ACB$

$x$ doesn't equal $50$

As a result, we only obtained one value of $x$ for which all requirements are met.

Therefore, the answer is $1$

#SPJ1