Lines and Angles Test

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Lines and Angles Test - 32

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Weekly Quiz Competition

Question 1
1 / -0

Find the complements of $$\displaystyle { 68 }^{ o }$$ and $$\displaystyle { 33 }^{ o }$$ respectively.

A
$$22^o, 57^o$$

B
$$57^o, 22^o$$

C
$$33^o, 67^o$$

D
$$67^o, 33^o$$

Solution

We know, two angles are complementary when they add up to $$90^o.$$

Given, measure of one complementary angle is $$68^o.$$

$$\Rightarrow$$ Measure of other complementary angle $$=90^o-68^o$$ $$=22^o.$$

$$\therefore$$ Measure of a complementary angle of $$ 68^{o}=22^o$$.

Also given, measure of one complementary angle is $$33^o.$$
$$\Rightarrow$$ Measure of other complementary angle $$=90^o-33^o$$ $$=57^o.$$

$$\therefore$$ Measure of a complementary angle of $$ 33^{o}=57^o$$.

Hence, option $$A$$ is correct.
Question 2
1 / -0

In $$\Delta PQR, \angle Q = 23^{o}$$ and $$\angle R = 20^{o}.$$ What is the measure of $$\angle P?$$

A
$$134^{o}$$

B
$$135^{o}$$

C
$$136^{o}$$

D
$$137^{o}$$

Solution

By angle sum property, the sum of angles is $$180^o$$.
$$\angle P+ \angle Q + \angle R = 180$$
$$\angle P + 23^{o} + 20^{o} = 180^{o}$$
$$\angle P = 180^{o} - 43^{o}$$
$$\angle P = 137^{o}.$$
Question 3
1 / -0

Which of the following indicated figures are examples of adjacent angles.

A
(i) and (iv)

B
(i) and (ii)

C
(i) and (iii)

D
(ii) and (iv)

Solution

(i) and (iv) have common side, common vertex and common arm. So, these are adjacent.
Question 4
1 / -0

In $$\Delta ABC, \angle A = 56^{o}$$ and $$\angle B = 60^{o}.$$ What is the measure of $$\angle C?$$

A
$$64^{o}$$

B
$$65^{o}$$

C
$$66^{o}$$

D
$$67^{o}$$

Solution

By angle sum property, the sum of angles is $$180^o$$.
$$\angle A+ \angle B + \angle C = 180^{o}$$
$$56^{o} + 60^{o} + \angle C = 180^{o}$$
$$\angle C = 180^{o} - 116^{o}$$
$$\angle C = 64^{o}$$
Question 5
1 / -0

In $$\Delta ABC,$$ $$\angle A = 43^{o}$$ and $$\angle C = 70^{o}.$$ What is the measure of $$\angle B?$$

A
$$63^{o}$$

B
$$65^{o}$$

C
$$66^{o}$$

D
$$67^{o}$$

Solution

By angle sum property, the sum of angles is $$180^o$$.
$$\angle A+ \angle B + \angle C = 180^{o}$$
$$43^{o} +\angle B + 70^{o} = 180^{o}$$
$$\angle B = 180^{o} - 113^{o}$$
$$\angle B = 67^{o}.$$
Question 6
1 / -0

If m$$\angle$$ $$XYZ = 86^o$$ and m$$\angle$$ $$XZY = 23^o$$. What is m$$\angle$$ $$YXZ$$ in the triangle?

A
$$70^o$$

B
$$71^o$$

C
$$72^o$$

D
$$73^o$$

Solution

By angle sum property, the sum of angles is $$180^o$$.
$$86^o + 23^o +$$ m$$\angle$$ $$YXZ = 180^o$$
m$$\angle$$ $$YXZ = 180^o - 109^o$$
m$$\angle$$ $$YXZ = 71 ^o$$.
Question 7
1 / -0

Find the measure of an exterior angle at the base of an isosceles triangle whose vertex angle measures $$50^o$$.

A
$$100^o$$

B
$$110^o$$

C
$$115^o$$

D
$$65^o$$

Solution

The $$2$$ base angles of an isosceles triangle are equal, so we'll represent each as $$x$$.

Now, by using the angle sum property,
$$x + x + 50^o = 180^o$$
$$\Rightarrow 2x = 180^o - 50^o$$
$$\Rightarrow 2x = 130^o$$
$$\Rightarrow x = 65^o$$

So, exterior angle is equal to the sum of the two non-adjacent interior angles.
Therefore,
$$y = x + 50^o$$
$$y = 50^o + 65^o$$
$$ = 115^o$$

Hence, option $$C$$ is correct.
Question 8
1 / -0

In quadrilateral $$PQRS$$ above, for what value of $$x$$, $$PS\parallel QR$$?

A
$$158$$

B
$$132$$

C
$$120$$

D
$$110$$

E
$$70$$

Solution

It is given that $$PS || QR$$.
Also, $$\angle SPQ=70^{0}$$

$$PQ$$ acts as a transversal since $$PS||QR$$.

Hence, $$\angle SPQ +\angle RQP=180^{0}$$

$$\Rightarrow 70^{0}+\angle Q=180^{0}$$

$$\Rightarrow \angle Q=110^{0}$$

Hence, the value of $$x$$ is $$110$$.
Question 9
1 / -0

In the figure, if $$\angle q=140$$, what is the value of $$\angle r- \angle p$$?

A
$$0^o$$

B
$$10^o$$

C
$$90^o$$

D
$$130^o$$

Solution

Angles $$p$$ and $$q$$ form a linear pair as seen in the figure.
Since $$\angle q = 140^o, \angle p = 180^o-40^o=40^o$$.

Also, the angle vertically opposite to $$p$$ in the right angled triangle will also be equal to $$p=40^o$$.
Since $$\angle r $$ is the exterior angle of the given right angled triangle, $$r = 90^o + 40^o = 130^o$$
$$\therefore \angle r - \angle p = 90^o$$.

Hence, option $$C$$ is correct.
Question 10
1 / -0

If m$$\angle$$ $$PRQ = 45^o$$ and m$$\angle$$ $$QPR = 68^o$$. What is m$$\angle$$ $$PQR$$ in the triangle?

A
$$67^o$$

B
$$68^o$$

C
$$69^o$$

D
$$70^o$$

Solution

by angle sum property, the sum of angles is $$180^o$$.

$$m\angle PRQ + m\angle QPR + m\angle PQR$$ $$= 180^o$$
$$45^o + 68^o +$$ m $$\angle$$ $$PQR = 180^o$$
m$$\angle$$ $$PQR = 180^o - 113^o$$
m$$\angle$$ $$PQR = 67^o$$ .