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In the diagram, QTR is a straight line and In the diagram, QTR is a straight line and o. find the sin of A. (frac{8}{15}) B. (frac{2}{3}) C. (frac{3}{4}) D. (frac{15}{16}) Correct Answer: Option C Explanation (frac{10}{sin 30^o} = frac{15}{sin x} = frac{10}{0.5} = frac{15}{sin x}) (frac{15}{20} = sin x) sin x = (frac{15}{20} = frac{3}{4}) N.B x = `, props: [‘comment’], data() { return {} }, methods: {} }) {{ reply.posted_by.display_name }}:   {{ reply.created_at }} `, props: [‘comment’, ‘reply’, ‘rate_selection’, ‘token’], data() { return {} }, methods: {} }) {{ comment.posted_by.display_name }}{{ comment.created_at }}

In the diagram, QTR is a straight line and o. find the sin of A. (frac{8}{15}) B. (frac{2}{3})…
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TQ is tangent to circle XYTR, TQ is tangent to circle XYTR, o, RTQ = 40o. fIND A. 108o B. 121o C. 140o D. 148o Correct Answer: Option A Explanation o (alternate segment)o(Angles in the same segments)o + 32o = 72oo(Supplementary) 72o + oo – 72o = 108o `, props: [‘comment’], data() { return {} }, methods: {} }) {{ reply.posted_by.display_name }}:   {{ reply.created_at }} `, props: [‘comment’, ‘reply’, ‘rate_selection’, ‘token’], data() { return {} }, methods: {} })

TQ is tangent to circle XYTR, o, RTQ = 40o. fIND A. 108o B. 121o C. 140o D.…
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