9514 1404 393
Answer:
(i) x° = 70°, y° = 20°
(ii) ∠BAC ≈ 50.2°
(iii) 120
(iv) 300
Step-by-step explanation:
(i) Angle x° is congruent with the one marked 70°, as they are "alternate interior angles" with respect to the parallel north-south lines and transversal AB.
x = 70
The angle marked y° is the supplement to the one marked 160°.
y = 20
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(ii) The triangle interior angle at B is x° +y° = 70° +20° = 90°, so triangle ABC is a right triangle. With respect to angle BAC, side BA is adjacent, and side BC is opposite. Then ...
tan(∠BAC) = BC/BA = 120/100 = 1.2
∠BAC = arctan(1.2) ≈ 50.2°
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(iii) The bearing of C from A is the sum of the bearing of B from A and angle BAC.
bearing of C = 70° +50.2° = 120.2°
The three-digit bearing of C from A is 120.
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(iv) The bearing of A from C is 180 added to the bearing of C from A:
120 +180 = 300
The three-digit bearing of A from C is 300.
2:3:4 gives 9 equal parts.
1.20 divided by 9 is 0.133.
2 * 0.133 would be 0.266m
3 * 0.133 would be 0.399m
4 * 0.133 would be 0.533m
To check this calculate 0.533 + 0.399 + 0.266 would be 1.198m (close enough)
Given:
Tangent segment MN = 6
External segment NQ = 4
Secant segment NP =x + 4
To find:
The length of line segment PQ.
Solution:
Property of tangent and secant segment:
If a secant and a tangent intersect outside a circle, then the product of the secant segment and external segment is equal to the product of the tangent segment.



Subtract 16 from both sides.


Divide by 4 on both sides.


The length of line segment PQ is 5 units.
we know that
the two triangles in the figure are similar, because they have three equal angles
therefore
The ratio of the corresponding sides are equal

therefore
<u>the answer is the option C</u>

G >\ -4 ( g is greater than or equal to negative 4 )