Usaid M.

asked • 02/09/19

Please help me solve this question. The figure shows a triangle ABC where ˂ABC = 35°. H lies on BC such that the length of HC is twice that of BH. Find <ACB.



I did apple triangle bisector theorem and then laws of sines, but the answer wasn't the same as textbook. The correct answer is 19.3°

Doug C.

Your question must be missing a key piece of information? It is possible to construct infinitely many triangles satisfying the given conditions. In each of those triangles the measure of angle C is different. Could the figure also give you the length of AC or AB?
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02/09/19

Philip P.

Is AH an angle bisector? Are the lengths of any sides given?
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02/09/19

Usaid M.

Nope no lengths were given, that is why I am confused, the text book just says HC is twice the size of BH. And yes AH is an angle bisector. This question is one of the advanced level questions in the chapter of Trignometrical Ratios, but I am sure the Law of sine applies here
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02/09/19

Usaid M.

And could you please help me with it? I am stuck on this question for 2 days, it is the last question of my exercise.
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02/09/19

Usaid M.

And the triangle ABC is a non right angled triangle but forms right angled triangle when divided into triangle ABH and TRIANGLE ACH
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02/09/19

Usaid M.

And it is a Seventh Grade question so I don't think the solution would include very core and complex mathematical theologies.
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02/09/19

Usaid M.

The way I did was by using triangle director theorem, I found AB is twice that of AC, SO AB = × AND AC = 2×. THEN I DID APPLY THE RULE OF SINE ANF AFTER ALL THE MATHS I GOT THE ANSWER OF 17.5° BUT THE ANSWER IS 19.3°
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02/09/19

1 Expert Answer

By:

KATHERINE C. answered • 02/09/19

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if 2 right angled triangles are formed, you can use tangent of the angles to solve it

tan (35) = AH/BH

tan(<ACH) = AH/HC = AH/2BH

this means

tan(35)* BH = tan(<ACH) * 2BH

if you solve it, <ACH will be 19.3 degrees

(note <ACH is the same as <ACB)

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Usaid M.

Thanks for helping me again. You are just great. Literally for helping me. My two days of tension would finally go away. Thank you so much, and sorry if it appears to be informal.
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02/09/19

Philip P.

Part of the problem, Usaid, is that you did not provide all of the information needed to solve the problem. The fact that triangles ABH and ACH are right triangles is information needed to solve the problem, but which you left out. If you want help, please be sure to provide ALL of the information in the problem.
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02/09/19

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